Chapter 6: Application of Derivatives

The area $$A$$ of a circle of radius $$r$$ is given by $$A = \pi r^2$$.

The rate of change of the area with respect to the radius is the derivative $$\dfrac{dA}{dr}$$.

$$\dfrac{dA}{dr} = \dfrac{d}{dr}\left(\pi r^2\right) = 2\pi r$$

When $$r = 5$$ cm,

$$\dfrac{dA}{dr}\bigg|_{r=5} = 2\pi(5) = 10\pi$$

Hence the area of the circle is changing at the rate of $$10\pi\ \mathrm{cm^2/cm}$$ when $$r = 5$$ cm.

Let the edge of the cube be $$x$$ cm at time $$t$$ seconds. Then the volume is $$V = x^3$$ and the surface area is $$S = 6x^2$$.

We are given $$\dfrac{dV}{dt} = 9\ \mathrm{cm^3/s}$$.

Differentiating $$V = x^3$$ with respect to $$t$$:

$$\dfrac{dV}{dt} = 3x^2\dfrac{dx}{dt} \quad\Rightarrow\quad 9 = 3x^2\dfrac{dx}{dt} \quad\Rightarrow\quad \dfrac{dx}{dt} = \dfrac{3}{x^2}$$

Now differentiate $$S = 6x^2$$ with respect to $$t$$:

$$\dfrac{dS}{dt} = 12x\dfrac{dx}{dt} = 12x\cdot\dfrac{3}{x^2} = \dfrac{36}{x}$$

When $$x = 10$$ cm,

$$\dfrac{dS}{dt} = \dfrac{36}{10} = 3.6$$

Hence the surface area is increasing at the rate of $$3.6\ \mathrm{cm^2/s}$$.

Let $$r$$ be the radius of the circular wave and $$A$$ the area enclosed by it at time $$t$$ seconds. Then $$A = \pi r^2$$.

The waves move at $$4$$ cm/s, so the radius increases at $$\dfrac{dr}{dt} = 4\ \mathrm{cm/s}$$.

Differentiating $$A = \pi r^2$$ with respect to $$t$$:

$$\dfrac{dA}{dt} = 2\pi r\dfrac{dr}{dt}$$

When $$r = 10$$ cm,

$$\dfrac{dA}{dt} = 2\pi(10)(4) = 80\pi$$

Hence the enclosed area is increasing at the rate of $$80\pi\ \mathrm{cm^2/s}$$.

Since the length is decreasing, $$\dfrac{dx}{dt} = -3$$ cm/min, and since the width is increasing, $$\dfrac{dy}{dt} = 2$$ cm/min. At the instant of interest, $$x = 10$$ cm and $$y = 6$$ cm.

(a) Perimeter. The perimeter is $$P = 2(x + y)$$.

$$\dfrac{dP}{dt} = 2\left(\dfrac{dx}{dt} + \dfrac{dy}{dt}\right) = 2(-3 + 2) = -2$$

Hence the perimeter is decreasing at the rate of $$2$$ cm/min.

(b) Area. The area is $$A = xy$$. By the product rule,

$$\dfrac{dA}{dt} = x\dfrac{dy}{dt} + y\dfrac{dx}{dt} = 10(2) + 6(-3) = 20 - 18 = 2$$

Hence the area is increasing at the rate of $$2\ \mathrm{cm^2/min}$$.

The marginal cost (MC) is the instantaneous rate of change of total cost with respect to the level of output, i.e. $$\mathrm{MC} = \dfrac{dC}{dx}$$.

$$\dfrac{dC}{dx} = \dfrac{d}{dx}\left(0.005x^3 - 0.02x^2 + 30x + 5000\right) = 0.015x^2 - 0.04x + 30$$

When $$x = 3$$ units are produced,

$$\mathrm{MC} = 0.015(3)^2 - 0.04(3) + 30 = 0.015(9) - 0.12 + 30$$

$$\mathrm{MC} = 0.135 - 0.12 + 30 = 30.015$$

Hence the marginal cost when 3 units are produced is Rs $$30.015$$, i.e. about Rs $$30.02$$.

The marginal revenue (MR) is the rate of change of total revenue with respect to the number of units sold, i.e. $$\mathrm{MR} = \dfrac{dR}{dx}$$.

$$\dfrac{dR}{dx} = \dfrac{d}{dx}\left(3x^2 + 36x + 5\right) = 6x + 36$$

When $$x = 5$$,

$$\mathrm{MR} = 6(5) + 36 = 30 + 36 = 66$$

Hence the marginal revenue when $$x = 5$$ is Rs $$66$$.

Let the length of an edge of the cube at any instant be $$x\,\text{cm}$$.

Volume: $$V = x^{3}$$ Surface area: $$S = 6x^{2}$$

Given rate of change of volume: $$\dfrac{dV}{dt}=8\;\text{cm}^{3}\!/\text{s}$$

Required: $$\dfrac{dS}{dt}$$ when $$x = 12\,\text{cm}$$.

1. Relate $$\dfrac{dx}{dt}$$ to $$\dfrac{dV}{dt}$$

Differentiate $$V = x^{3}$$ with respect to $$t$$:

$$\dfrac{dV}{dt} = 3x^{2}\dfrac{dx}{dt}$$

\(\Rightarrow\) $$\dfrac{dx}{dt}=\dfrac{\dfrac{dV}{dt}}{3x^{2}}$$

2. Differentiate the surface area

$$S = 6x^{2}\;\Rightarrow\;\dfrac{dS}{dt}=12x\dfrac{dx}{dt}$$

3. Substitute $$\dfrac{dx}{dt}$$

$$\dfrac{dS}{dt}=12x\left(\dfrac{\dfrac{dV}{dt}}{3x^{2}}\right)=\dfrac{12x}{3x^{2}}\dfrac{dV}{dt}=\dfrac{4}{x}\dfrac{dV}{dt}$$

4. Evaluate at $$x = 12\,\text{cm}$$

$$\dfrac{dS}{dt}=\dfrac{4}{12}\times 8=\dfrac{1}{3}\times 8=\dfrac{8}{3}$$

Units: $$\text{cm}^{2}\!/\text{s}$$.

Therefore, when the edge length is 12 cm, the surface area is increasing at $$\dfrac{8}{3}\,\text{cm}^{2}\!/\text{s}$$.

Let $$r\,(\text{cm})$$ denote the radius and $$t\,(\text{s})$$ denote time. The area of the circle is

$$A = \pi r^{2}.$$

Differentiate both sides with respect to time $$t$$ (using the chain rule):

$$\frac{dA}{dt} = 2\pi r\,\frac{dr}{dt}.$$

We are given

$$\frac{dr}{dt} = 3\,\text{cm/s}, \qquad r = 10\,\text{cm}.$$

Substituting these values,

$$\frac{dA}{dt} = 2\pi(10)(3) = 60\pi.$$

Hence, the area is increasing at the rate of $$60\pi\,\text{cm}^{2}\text{/s}$$ when the radius is $$10\,\text{cm}$$.

Let $$x\,(\text{cm})$$ be the length of an edge of the cube at time $$t\,(\text{s})$$.

The edge is increasing at the rate

$$\frac{dx}{dt}=3\ \text{cm/s}.$$

The volume $$V$$ of the cube is related to $$x$$ by

$$V = x^3.$$

Differentiating both sides with respect to $$t$$ by the chain rule,

$$\frac{dV}{dt}=3x^2\frac{dx}{dt}.$$

At the instant when $$x = 10\ \text{cm}$$, substitute $$x = 10$$ and $$\frac{dx}{dt}=3$$:

$$\frac{dV}{dt}=3(10)^2(3)=3\times100\times3=900\ \text{cm}^3\!/\!\text{s}.$$

Therefore, the volume of the cube is increasing at $$900\ \text{cm}^3\,\text{s}^{-1}$$ when the edge is $$10\ \text{cm}$$ long.

Given data

• The circular wave front has radius $$r\;\text{cm}$$.
• The radius is increasing at a constant rate $$\dfrac{dr}{dt}=5\;\text{cm\,/\,s}$$ (because the waves move outwards at 5 cm s⁻¹).
• We are asked for the rate at which the enclosed area $$A$$ is increasing when $$r = 8\;\text{cm}$$, i.e. we want $$\dfrac{dA}{dt}\Big|_{r=8}$$.

Step 1 — Express the area in terms of the radius

The area of a circle of radius $$r$$ is

$$A = \pi r^{2}$$

Step 2 — Differentiate with respect to time $$t$$

Both $$A$$ and $$r$$ change with time, so apply the chain rule:

$$\frac{dA}{dt} = \frac{d}{dt}\,(\pi r^{2}) = 2\pi r\,\frac{dr}{dt}$$

Step 3 — Substitute the known values

At the instant when $$r = 8\;\text{cm}$$ and $$\dfrac{dr}{dt} = 5\;\text{cm\,/\,s}$$,

$$\frac{dA}{dt} = 2\pi(8)\,(5)$$

$$= 80\pi$$

Step 4 — Attach the correct units

Because area is measured in square centimetres and time in seconds,

$$\frac{dA}{dt} = 80\pi\;\text{cm}^{2}\!\,\text{s}^{-1}$$

Thus, the enclosed area is increasing at a rate of $$80\pi\;\text{cm}^{2}\,\text{s}^{-1}$$ when the radius equals 8 cm.

Given data

• Let the radius of the circle at time t be $$r \text{ cm}$$.
• The radius is increasing at the rate $$\dfrac{dr}{dt}=0.7\;\text{cm s}^{-1}$$.

Quantity to be found

The rate of increase of the circumference $$C$$ of the circle, i.e. $$\dfrac{dC}{dt}$$.

Step 1. Relate the variables

The circumference of a circle of radius $$r$$ is

$$C = 2\pi r$$.

Step 2. Differentiate with respect to time t

Both $$C$$ and $$r$$ are functions of time, so apply the chain rule:

$$\frac{dC}{dt}=\frac{d}{dt}(2\pi r)=2\pi\frac{dr}{dt}.$$

Step 3. Substitute the known value of $$\dfrac{dr}{dt}$$

$$\frac{dC}{dt}=2\pi\,(0.7)=1.4\pi\;\text{cm s}^{-1}.$$

Step 4. Numerical value (optional)

Using $$\pi\approx3.14$$,

$$\frac{dC}{dt}\approx1.4\times3.14\approx4.40\;\text{cm s}^{-1}.$$

Conclusion

The circumference is increasing at $$1.4\pi\,\text{cm s}^{-1}\,(\approx4.40\,\text{cm s}^{-1}).$$

Let the length and width of the rectangle at time $$t$$ be $$x(t)$$ and $$y(t)$$ respectively.

Given rates of change:

• $$\dfrac{dx}{dt} = -5\,\text{cm/min}$$  (length is decreasing)
• $$\dfrac{dy}{dt} = +4\,\text{cm/min}$$  (width is increasing)

(a) Rate of change of the perimeter

The perimeter $$P$$ of a rectangle is $$P = 2(x + y)$$.

Differentiating with respect to time $$t$$:

$$\dfrac{dP}{dt} = 2\bigl(\dfrac{dx}{dt} + \dfrac{dy}{dt}\bigr).$$

Substituting $$\dfrac{dx}{dt} = -5$$ and $$\dfrac{dy}{dt} = 4$$,

$$\dfrac{dP}{dt} = 2(-5 + 4) = 2(-1) = -2\,\text{cm/min}.$$

(b) Rate of change of the area

The area $$A$$ is $$A = x y$$.

Differentiating with respect to time:

$$\dfrac{dA}{dt} = x\,\dfrac{dy}{dt} + y\,\dfrac{dx}{dt}$$  (product rule).

At the instant when $$x = 8\,\text{cm}$$ and $$y = 6\,\text{cm}$$,

$$\dfrac{dA}{dt} = 8(4) + 6(-5) = 32 - 30 = 2\,\text{cm}^2/\text{min}.$$

Therefore, when $$x = 8\,\text{cm}$$ and $$y = 6\,\text{cm}$$:

• The perimeter is decreasing at $$2\,\text{cm/min}$$.
• The area is increasing at $$2\,\text{cm}^2/\text{min}$$.

Given data

The balloon remains spherical, so its volume at any instant is $$V=\frac{4}{3}\pi r^{3}$$.
Gas is being pumped in at the rate $$\frac{dV}{dt}=900\,\text{cm}^{3}\text{s}^{-1}$$.

We have to find $$\frac{dr}{dt}$$ when $$r=15\,\text{cm}$$.

Step 1 : Differentiate the volume

Differentiate $$V=\frac{4}{3}\pi r^{3}$$ with respect to time $$t$$:

$$\frac{dV}{dt}=\frac{d}{dt}\left(\frac{4}{3}\pi r^{3}\right)=4\pi r^{2}\frac{dr}{dt}$$  (by the chain rule).

Step 2 : Insert the given values

Substituting $$\frac{dV}{dt}=900$$ and $$r=15$$:

$$900 = 4\pi (15)^{2}\frac{dr}{dt}$$

Step 3 : Solve for $$\frac{dr}{dt}$$

$$\frac{dr}{dt}=\frac{900}{4\pi\times225}=\frac{900}{900\pi}=\frac{1}{\pi}\;\text{cm s}^{-1}$$

Conclusion

When the radius of the balloon is $$15\,\text{cm}$$, it is increasing at the rate $$\boxed{\dfrac{1}{\pi}\,\text{cm s}^{-1}}$$.

Let $$r\text{ cm}$$ denote the radius of the spherical balloon and $$V\text{ cm}^3$$ its volume.

The volume of a sphere is given by $$V = \frac{4}{3}\pi r^{3}.$$

Differentiating with respect to $$r$$ gives the rate of change of volume with the radius:

$$\frac{\mathrm dV}{\mathrm dr} = \frac{\mathrm d}{\mathrm dr}\Bigl(\frac{4}{3}\pi r^{3}\Bigr) = \frac{4}{3}\pi \cdot 3r^{2} = 4\pi r^{2}.$$

When $$r = 10\text{ cm}$$,

$$\left.\frac{\mathrm dV}{\mathrm dr}\right|_{r = 10} = 4\pi (10)^{2} = 4\pi \times 100 = 400\pi.$$

Therefore, the volume is increasing with the radius at the rate of $$400\pi\text{ cm}^{2}$$ when the radius is $$10\text{ cm}$$.

Given data

• Length of the ladder (constant): $$5\text{ m}$$.
• Let $$x(t)$$ be the horizontal distance (in metres) of the foot of the ladder from the wall.
• Let $$y(t)$$ be the vertical distance (in metres) of the top of the ladder from the ground (its height on the wall).
• The foot of the ladder is being pulled away from the wall at the rate $$\displaystyle\frac{dx}{dt}=2\text{ cm s}^{-1}=0.02\text{ m s}^{-1}$$.
• We need $$\displaystyle\frac{dy}{dt}$$ when $$x=4\text{ m}$$.

1. Geometric relation between $$x$$ and $$y$$

The ladder, the wall and the ground form a right-angled triangle. By Pythagoras’ theorem,

$$x^{2}+y^{2}=5^{2}.$$

2. Differentiate with respect to time $$t$$

Because the length of the ladder is constant, we differentiate the above relation implicitly:

$$\frac{d}{dt}\,(x^{2}+y^{2})=\frac{d}{dt}\,(25).$$

This gives

$$2x\,\frac{dx}{dt}+2y\,\frac{dy}{dt}=0.$$

3. Solve for $$\displaystyle\frac{dy}{dt}$$

First cancel the common factor $$2$$:

$$x\,\frac{dx}{dt}+y\,\frac{dy}{dt}=0.$$

Re-arrange for $$\displaystyle\frac{dy}{dt}$$:

$$y\,\frac{dy}{dt}=-x\,\frac{dx}{dt}\;\;\Longrightarrow\;\;\frac{dy}{dt}=-\frac{x}{y}\,\frac{dx}{dt}.$$

4. Evaluate at the instant when $$x=4\text{ m}$$

First compute $$y$$ using $$x^{2}+y^{2}=25$$:

$$y^{2}=25-4^{2}=25-16=9\;\;\Longrightarrow\;\;y=3\text{ m}$$ (positive height).

Now substitute $$x=4$$, $$y=3$$ and $$\displaystyle\frac{dx}{dt}=0.02\text{ m s}^{-1}$$:

$$\frac{dy}{dt}=-\frac{4}{3}\times 0.02\;\text{m s}^{-1}.$$

Compute the product:

$$\frac{dy}{dt}= -0.026666\ldots\;\text{m s}^{-1}.$$

5. Convert back to centimetres per second

Since $$1\text{ m}=100\text{ cm}$$,

$$-0.026666\ldots\text{ m s}^{-1}= -0.026666\ldots\times100\text{ cm s}^{-1}= -2.6666\ldots\text{ cm s}^{-1}.$$

Exactly,

$$\frac{dy}{dt}= -\frac{8}{3}\text{ cm s}^{-1}.$$

Interpretation

The negative sign shows that the height of the ladder on the wall is decreasing. Its magnitude is $$\tfrac83\text{ cm s}^{-1}\;\approx\;2.67\text{ cm s}^{-1}$$.

The particle moves on the curve $$6y = x^3 + 2$$, so both $$x$$ and $$y$$ depend on time $$t$$.

We are told that, at the required points, the rate of change of $$y$$ is eight times the rate of change of $$x$$, i.e.

$$\frac{dy}{dt} = 8\,\frac{dx}{dt}.$$

By the chain rule, $$\dfrac{dy}{dt} = \dfrac{dy}{dx}\,\dfrac{dx}{dt}$$. Substitute this in the above relation:

$$\dfrac{dy}{dx}\,\dfrac{dx}{dt} = 8\,\dfrac{dx}{dt}.$$

If $$\dfrac{dx}{dt} \neq 0$$, we may cancel it to obtain

$$\dfrac{dy}{dx} = 8.$$

Thus we need the points on the curve where the slope (derivative) equals $$8$$.

Differentiating the given relation with respect to $$x$$:

$$6\,\dfrac{dy}{dx} = 3x^2 \quad\Longrightarrow\quad \dfrac{dy}{dx} = \dfrac{3x^2}{6} = \dfrac{x^2}{2}.$$

Set this equal to $$8$$:

$$\dfrac{x^2}{2} = 8 \;\Longrightarrow\; x^2 = 16 \;\Longrightarrow\; x = \pm 4.$$

Find the corresponding $$y$$-coordinates from the original curve:

• For $$x = 4$$: $$y = \dfrac{4^3 + 2}{6} = \dfrac{64 + 2}{6} = \dfrac{66}{6} = 11.$$
• For $$x = -4$$: $$y = \dfrac{(-4)^3 + 2}{6} = \dfrac{-64 + 2}{6} = \dfrac{-62}{6} = -\dfrac{31}{3}.$$

Hence, the particle’s $$y$$-coordinate changes eight times as fast as its $$x$$-coordinate at the points $$\bigl(4, 11\bigr)$$ and $$\bigl(-4, -\tfrac{31}{3}\bigr)$$.

Let the radius of the air bubble at time $$t$$ seconds be $$r$$ cm and its volume be $$V$$ cm³.

For a sphere

$$V = \frac{4}{3}\pi r^{3}$$

Differentiate both sides with respect to time $$t$$ (using the chain rule):

$$\frac{dV}{dt} = \frac{d}{dt}\Bigl(\tfrac{4}{3}\pi r^{3}\Bigr) = \frac{4}{3}\pi \cdot 3r^{2}\frac{dr}{dt} = 4\pi r^{2}\frac{dr}{dt}$$

The rate of change of radius is given:

$$\frac{dr}{dt} = \tfrac{1}{2}\text{ cm/s}$$

We need $$\dfrac{dV}{dt}$$ when $$r = 1\text{ cm}$$.

Substitute $$r = 1$$ and $$\dfrac{dr}{dt} = \tfrac{1}{2}$$ in $$\dfrac{dV}{dt} = 4\pi r^{2}\dfrac{dr}{dt}$$:

$$\frac{dV}{dt} = 4\pi (1)^{2}\left(\tfrac{1}{2}\right) = 4\pi\times\tfrac{1}{2} = 2\pi$$

Hence the volume of the bubble is increasing at the rate of $$2\pi\text{ cm}^{3}\!\big/\text{s}$$ when the radius is $$1\text{ cm}$$.

A spherical balloon of radius $$r$$ has volume

$$V=\frac{4}{3}\,\bigl(\pi r^{3}\bigr).$$

The diameter varies with $$x$$ as

$$D=\frac{3}{2}(2x+1).$$

Hence the radius is half of the diameter:

$$r=\frac{D}{2}=\frac{1}{2}\Bigl[\frac{3}{2}(2x+1)\Bigr]=\frac{3}{2}x+\frac{3}{4}=\frac{6x+3}{4}.$$

Express the volume in terms of $$x$$:

$$V(x)=\frac{4}{3}\,\pi\Bigl(\frac{3}{2}x+\frac{3}{4}\Bigr)^{3}.$$

Differentiate using the chain rule.

First compute the rate of change of radius:

$$\frac{\mathrm d r}{\mathrm d x}=\frac{3}{2}.$$

Next relate the derivatives:

$$\frac{\mathrm d V}{\mathrm d x}=\frac{\mathrm d V}{\mathrm d r}\,\frac{\mathrm d r}{\mathrm d x}.$$

Because $$\dfrac{\mathrm d V}{\mathrm d r}=4\pi r^{2}$$, we obtain

$$\frac{\mathrm d V}{\mathrm d x}=4\pi r^{2}\,\frac{3}{2}=6\pi r^{2}.$$

Substitute $$r=\dfrac{6x+3}{4}$$:

$$\frac{\mathrm d V}{\mathrm d x}=6\pi\Bigl(\frac{6x+3}{4}\Bigr)^{2}=\frac{6\pi}{16}(6x+3)^{2}=\frac{3\pi}{8}\,9(2x+1)^{2}.$$

Simplifying,

$$\frac{\mathrm d V}{\mathrm d x}=\frac{27\pi}{8}(2x+1)^{2}.$$

Thus, the required rate of change of the balloon’s volume with respect to $$x$$ is

$$\boxed{\dfrac{\mathrm d V}{\mathrm d x}=\dfrac{27\pi}{8}(2x+1)^{2}}.$$

Given data

• Rate at which sand is poured (rate of change of volume): $$\dfrac{dV}{dt}=12\;\text{cm}^3\!\text{/s}$$.
• The sand forms a right circular cone for which the height is always one–sixth of its base-radius. Hence
  $$h=\dfrac{r}{6}\;\Longrightarrow\;r=6h$$.
• Required: $$\dfrac{dh}{dt}$$ when $$h=4\;\text{cm}$$.

1. Express the volume solely in terms of h

For a right circular cone, the volume is

$$V=\tfrac13\pi r^{2}h.$$

Substitute $$r=6h$$:

$$V=\tfrac13\pi(6h)^{2}h=\tfrac13\pi\cdot36h^{3}=12\pi h^{3}.$$

2. Differentiate with respect to t

Because $$V$$ and $$h$$ vary with time, apply the chain rule:

$$\dfrac{dV}{dt}=\dfrac{d}{dt}\bigl(12\pi h^{3}\bigr)=12\pi\cdot3h^{2}\dfrac{dh}{dt}=36\pi h^{2}\dfrac{dh}{dt}.$$

3. Insert the known numerical values

At the instant when $$h=4\,\text{cm}$$, we have

$$12=36\pi(4)^{2}\dfrac{dh}{dt}$$

$$\Longrightarrow\;12=36\pi\cdot16\,\dfrac{dh}{dt}=576\pi\,\dfrac{dh}{dt}.$$

4. Solve for  $$\dfrac{dh}{dt}$$

$$\dfrac{dh}{dt}=\dfrac{12}{576\pi}=\dfrac{1}{48\pi}\;\text{cm/s}$$.

Numerically, $$\dfrac{1}{48\pi}\approx0.00663\,\text{cm/s}.$$

Hence, the height of the sand cone is increasing at $$\boxed{\dfrac{1}{48\pi}\;\text{cm s}^{-1}}$$ when its height is 4 cm.

The total cost function is

$$C(x)=0.007x^{3}-0.003x^{2}+15x+4000$$

Step 1 — Find the marginal-cost function.
The marginal cost is the first derivative of the cost function with respect to $$x$$:

$$C'(x)=\frac{d}{dx}\,(0.007x^{3}-0.003x^{2}+15x+4000).$$

Differentiate term by term:

\[\begin{aligned}C'(x)&=0.007\times3x^{2}-0.003\times2x+15\\ &=0.021x^{2}-0.006x+15.\end{aligned}\]

Step 2 — Evaluate at $$x=17$$ units.

$$C'(17)=0.021\,(17)^{2}-0.006\,(17)+15.$$

Compute each term:

$$17^{2}=289,$$

$$0.021\times289=6.069,$$

$$0.006\times17=0.102.$$

Therefore

\[\begin{aligned}C'(17)&=6.069-0.102+15\\ &=5.967+15\\ &=20.967.\end{aligned}\]

Thus, the marginal cost when 17 units are produced is approximately Rs 20.97.

The marginal revenue is defined as the rate of change of revenue with respect to the number of units sold, that is, $$\frac{dR}{dx}$$.

Given $$R(x)=13x^2+26x+15$$.

Differentiate term by term:

$$\frac{dR}{dx}=\frac{d}{dx}(13x^2)+\frac{d}{dx}(26x)+\frac{d}{dx}(15)$$

Using the power and constant rules:

$$\frac{dR}{dx}=26x+26+0=26x+26$$

Hence the marginal revenue function is $$26x+26$$.

Evaluate it at $$x=7$$:

$$\frac{dR}{dx}\bigg|_{x=7}=26(7)+26=182+26=208$$

Therefore, the marginal revenue when 7 units are sold is Rs 208.

The function is $$f(x)=7x-3$$, defined for every real number, so its domain is $$\mathbf{R}$$.

Step 1 · Differentiate.
Using the rule $$\dfrac{\mathrm d}{\mathrm dx}(ax+b)=a$$ we get
$$f'(x)=7.$$

Step 2 · Sign of the derivative.
The constant $$7$$ is positive for every $$x\in\mathbf{R}$$, so
$$f'(x)>0\quad\text{for all }x\in\mathbf{R}.$$

Step 3 · Apply the increasing-function test.
A function whose derivative is positive at each point of an interval is strictly increasing on that interval. Because $$f'(x)>0$$ for all $$x$$ in the whole real line, $$f$$ is strictly increasing on $$\mathbf{R}$$.

Therefore the given function $$f(x)=7x-3$$ is increasing on the entire set of real numbers.

Step 1 : Write the given function

The function under consideration is $$f(x)=x^3-3x^2+4x$$, where $$x\in\mathbf{R}$$.

Step 2 : Recall the criterion for monotonicity

A differentiable function is increasing on an interval if and only if its derivative is non-negative throughout that interval, and it is strictly increasing if the derivative is >0 everywhere.

Step 3 : Differentiate

Using the standard rules $$\dfrac{d}{dx}(x^n)=nx^{n-1}$$ and $$\dfrac{d}{dx}(k\,x)=k$$, we get

$$f'(x)=\dfrac{d}{dx}(x^3)-\dfrac{d}{dx}(3x^2)+\dfrac{d}{dx}(4x)=3x^2-6x+4.$$

Step 4 : Examine the sign of the derivative

The expression $$f'(x)=3x^2-6x+4$$ is a quadratic in $$x$$. To decide its sign for all real $$x$$, compute its discriminant.

For a quadratic $$ax^2+bx+c$$, the discriminant is $$\Delta=b^2-4ac$$.
Here $$a=3$$, $$b=-6$$ and $$c=4$$, so

$$\Delta=(-6)^2-4\times3\times4=36-48=-12.$$

Step 5 : Interpret the discriminant

Because $$\Delta=-12<0$$, the quadratic $$3x^2-6x+4$$ has no real roots. Moreover, its leading coefficient $$a=3>0$$, so the parabola opens upward. A parabola with no real roots and opening upward stays entirely above the $$x$$-axis, hence

$$3x^2-6x+4>0 \text{ for every real } x.$$

Step 6 : Conclude about monotonicity of $$f$$

Since $$f'(x)>0$$ for all $$x\in\mathbf{R}$$, the derivative never changes sign and is never zero. Therefore the given function $$f(x)=x^3-3x^2+4x$$ is strictly increasing on the entire real line.

Hence, the function is increasing on $$\mathbf{R}$$.

To find where $$f(x) = 4x^3 - 6x^2 - 72x + 30$$ is increasing or decreasing, we examine the sign of its first derivative.

Step 1 — First derivative.

$$f'(x) = 12x^2 - 12x - 72 = 12(x^2 - x - 6) = 12(x - 3)(x + 2).$$

Step 2 — Critical points.

Set $$f'(x) = 0$$:

$$12(x - 3)(x + 2) = 0 \quad\Longrightarrow\quad x = -2 \ \text{or}\ x = 3.$$

These points divide the real line into three sub-intervals: $$(-\infty, -2)$$, $$(-2, 3)$$, and $$(3, \infty)$$.

Step 3 — Sign of the derivative on each sub-interval.

• On $$(-\infty, -2)$$, take the test point $$x = -3$$. Then $$x - 3 = -6 \lt 0$$ and $$x + 2 = -1 \lt 0$$, so $$f'(-3) = 12(-6)(-1) = 72 \gt 0$$. Hence $$f'(x) \gt 0$$ on $$(-\infty, -2)$$.
• On $$(-2, 3)$$, take the test point $$x = 0$$. Then $$x - 3 = -3 \lt 0$$ and $$x + 2 = 2 \gt 0$$, so $$f'(0) = 12(-3)(2) = -72 \lt 0$$. Hence $$f'(x) \lt 0$$ on $$(-2, 3)$$.
• On $$(3, \infty)$$, take the test point $$x = 4$$. Then $$x - 3 = 1 \gt 0$$ and $$x + 2 = 6 \gt 0$$, so $$f'(4) = 12(1)(6) = 72 \gt 0$$. Hence $$f'(x) \gt 0$$ on $$(3, \infty)$$.

Step 4 — Conclusion.

• (a) $$f$$ is increasing where $$f'(x) \gt 0$$, i.e. on $$(-\infty, -2) \cup (3, \infty)$$.
• (b) $$f$$ is decreasing where $$f'(x) \lt 0$$, i.e. on $$(-2, 3)$$.

Given: $$f(x) = \sin 3x$$ on $$\left[0, \dfrac{\pi}{2}\right]$$.

Step 1 — First derivative.

$$f'(x) = \dfrac{d}{dx}(\sin 3x) = 3\cos 3x.$$

Step 2 — Critical points in the interval.

Set $$f'(x) = 0$$:

$$3\cos 3x = 0 \quad\Longrightarrow\quad \cos 3x = 0.$$

This holds when $$3x = \dfrac{\pi}{2} + k\pi$$ for some integer $$k$$, i.e. $$x = \dfrac{\pi}{6} + \dfrac{k\pi}{3}$$.

Within $$\left[0, \dfrac{\pi}{2}\right]$$, the solutions are $$x = \dfrac{\pi}{6}$$ (for $$k = 0$$) and $$x = \dfrac{\pi}{2}$$ (for $$k = 1$$, the right end-point). The only interior critical point is therefore $$x = \dfrac{\pi}{6}$$, which splits the interval into $$\left[0, \dfrac{\pi}{6}\right]$$ and $$\left[\dfrac{\pi}{6}, \dfrac{\pi}{2}\right]$$.

Step 3 — Sign of $$f'(x)$$ on each sub-interval.

• For $$0 \le x \lt \dfrac{\pi}{6}$$, we have $$0 \le 3x \lt \dfrac{\pi}{2}$$, so $$\cos 3x \gt 0$$, hence $$f'(x) = 3\cos 3x \gt 0$$. Thus $$f$$ is increasing on $$\left[0, \dfrac{\pi}{6}\right]$$.
• For $$\dfrac{\pi}{6} \lt x \le \dfrac{\pi}{2}$$, we have $$\dfrac{\pi}{2} \lt 3x \le \dfrac{3\pi}{2}$$, so $$\cos 3x \lt 0$$, hence $$f'(x) = 3\cos 3x \lt 0$$. Thus $$f$$ is decreasing on $$\left[\dfrac{\pi}{6}, \dfrac{\pi}{2}\right]$$.

Step 4 — Conclusion.

• (a) $$f$$ is increasing on $$\left[0, \dfrac{\pi}{6}\right]$$.
• (b) $$f$$ is decreasing on $$\left[\dfrac{\pi}{6}, \dfrac{\pi}{2}\right]$$.

At $$x = \dfrac{\pi}{6}$$ the derivative is zero, and the function transitions from increasing to decreasing at this point.

We study the monotonicity of $$f(x) = \sin x + \cos x$$ on $$[0, 2\pi]$$.

Step 1 — First derivative.

$$f'(x) = \cos x - \sin x.$$

Step 2 — Critical points.

Set $$f'(x) = 0$$:

$$\cos x - \sin x = 0 \quad\Longrightarrow\quad \tan x = 1.$$

In $$[0, 2\pi]$$ the solutions are $$x = \dfrac{\pi}{4}$$ and $$x = \dfrac{5\pi}{4}$$. These split $$[0, 2\pi]$$ into three sub-intervals: $$\left(0, \dfrac{\pi}{4}\right)$$, $$\left(\dfrac{\pi}{4}, \dfrac{5\pi}{4}\right)$$, and $$\left(\dfrac{5\pi}{4}, 2\pi\right)$$.

Step 3 — Sign of $$f'(x)$$ on each sub-interval.

• On $$\left(0, \dfrac{\pi}{4}\right)$$, take the test point $$x = 0$$: $$f'(0) = \cos 0 - \sin 0 = 1 - 0 = 1 \gt 0$$. Hence $$f$$ is increasing on $$\left[0, \dfrac{\pi}{4}\right]$$.
• On $$\left(\dfrac{\pi}{4}, \dfrac{5\pi}{4}\right)$$, take the test point $$x = \pi$$: $$f'(\pi) = \cos \pi - \sin \pi = -1 - 0 = -1 \lt 0$$. Hence $$f$$ is decreasing on $$\left[\dfrac{\pi}{4}, \dfrac{5\pi}{4}\right]$$.
• On $$\left(\dfrac{5\pi}{4}, 2\pi\right)$$, take the test point $$x = \dfrac{3\pi}{2}$$: $$f'\!\left(\dfrac{3\pi}{2}\right) = \cos\dfrac{3\pi}{2} - \sin\dfrac{3\pi}{2} = 0 - (-1) = 1 \gt 0$$. Hence $$f$$ is increasing on $$\left[\dfrac{5\pi}{4}, 2\pi\right]$$.

Step 4 — Conclusion.

• $$f$$ is increasing on $$\left[0, \dfrac{\pi}{4}\right] \cup \left[\dfrac{5\pi}{4}, 2\pi\right]$$.
• $$f$$ is decreasing on $$\left[\dfrac{\pi}{4}, \dfrac{5\pi}{4}\right]$$.

Step 1 : Recall the definition of an increasing function

A real-valued function $$f$$ is said to be increasing on an interval if for any two numbers $$x_1,x_2$$ in that interval, $$x_1<x_2\;\Rightarrow\;f(x_1)<f(x_2)$$. A convenient sufficient condition from calculus is:

• If $$f$$ is differentiable on the interval and its derivative satisfies $$f'(x)>0$$ for every $$x$$ in that interval, then $$f$$ is increasing there.

Step 2 : Differentiate the given function

The function is $$f(x)=3x+17$$, which is a linear (polynomial) function and hence differentiable for every real number $$x$$.

Derivative:

$$f'(x)=\dfrac{d}{dx}\,(3x)+\dfrac{d}{dx}\,(17)=3+0=3$$

Step 3 : Analyse the sign of the derivative

For every $$x\in\mathbf{R}$$ we have $$f'(x)=3$$. Clearly

$$3>0\quad\text{for all }x\in\mathbf{R}.$$

Step 4 : Conclude monotonicity

Because the derivative is positive everywhere on $$\mathbf{R}$$, the sufficient condition stated in Step 1 applies. Therefore, $$f(x)=3x+17$$ is increasing on the entire real line $$\mathbf{R}$$.

Result: The function is increasing on $$\mathbf{R}$$.

Given function
$$f(x)=e^{2x},\;x\in\mathbf R$$

Step 1 – Find the first derivative
Using the chain rule, for $$u(x)=2x$$ we have $$\dfrac{d}{dx}\,e^{u}=e^{u}\,u'(x)$$. Hence

$$f'(x)=\dfrac{d}{dx}\,e^{2x}=e^{2x}\,\dfrac{d}{dx}(2x)=e^{2x}\cdot2=2e^{2x}.$$

Step 2 – Sign of the derivative
For every real number $$x$$, the exponential term satisfies $$e^{2x}>0$$ (because any real exponential is positive). Therefore

$$2e^{2x}>0\quad\text{for all }x\in\mathbf R.$$

Step 3 – Apply the monotonicity test
A standard result from differential calculus states: if $$f'(x)>0$$ for every $$x$$ in an interval, then $$f(x)$$ is strictly increasing on that interval. Since we have shown $$f'(x)=2e^{2x}>0$$ for all real $$x$$, the condition is satisfied on the entire real line.

Conclusion
$$f(x)=e^{2x}$$ is strictly increasing on $$\mathbf R$$.

Given
$$y = \log(1+x) - \frac{2x}{2+x}, \; x>-1$$
(Here $$\log$$ denotes the natural logarithm.)

Step 1 : Differentiate.
\[\begin{aligned} \frac{dy}{dx} &= \frac{d}{dx}\bigl(\log(1+x)\bigr)\;- \frac{d}{dx}\Bigl(\frac{2x}{2+x}\Bigr) \\[4pt] &= \frac{1}{1+x} - \frac{(2)(2+x) - 2x(1)}{(2+x)^2} \quad \bigl[\text{quotient rule}\bigr] \\[4pt] &= \frac{1}{1+x} - \frac{4+2x-2x}{(2+x)^2} \\[2pt] &= \frac{1}{1+x} - \frac{4}{(2+x)^2}. \end{aligned}\]
Step 2 : Put over a common denominator.
\[\begin{aligned} \frac{dy}{dx} &= \frac{(2+x)^2}{(1+x)(2+x)^2} \, - \, \frac{4(1+x)}{(1+x)(2+x)^2} \\[4pt] &= \frac{(2+x)^2 - 4(1+x)}{(1+x)(2+x)^2}. \end{aligned}\]
Step 3 : Simplify the numerator.
\[(2+x)^2 = x^2 + 4x + 4,\qquad 4(1+x)=4+4x.\] Hence
\[ (2+x)^2 - 4(1+x) = (x^2 + 4x + 4) - (4 + 4x) = x^2. \] Therefore
$$\frac{dy}{dx}=\frac{x^2}{(1+x)(2+x)^2}.$$
Step 4 : Sign of the derivative.

• For the given domain $$x>-1$$ we have $$1+x>0$$ and $$2+x>1>0\;\Rightarrow\;(2+x)^2>0$$.
• The numerator $$x^2\ge 0$$, becoming zero only at $$x=0$$.

Given  $$y =[x (x-2)]^{2}=(x^{2}-2x)^{2}$$

Step 1 : Differentiate $$y$$

Put  $$u = x^{2}-2x \\[3pt] y=u^{2}$$

Then  $$\dfrac{dy}{dx}=2u\,\dfrac{du}{dx}$$

and  $$\dfrac{du}{dx}=2x-2=2(x-1)$$

Hence

$$\dfrac{dy}{dx}=2(x^{2}-2x)\;2(x-1)=4x\,(x-2)\,(x-1)$$

Step 2 : Locate critical points

$$\dfrac{dy}{dx}=0\;\Longrightarrow\;4x(x-1)(x-2)=0$$

Critical points: $$x=0,\;1,\;2$$.

Step 3 : Sign of the derivative

Step 4 : Intervals of increase

The derivative is positive on $$(0,1)$$ and $$(2,\infty)$$. At the points $$x=0$$, $$1$$ and $$2$$ the derivative vanishes only momentarily, so the function is still non-decreasing there. Therefore 

$$y$$ is increasing on $$[0,1]\;\cup\;[2,\infty).$$

Given function
$$y(\theta)=\dfrac{4\sin\theta}{2+\cos\theta}-\theta,\qquad\theta\in\left[0,\dfrac{\pi}{2}\right].$$

To prove that $$y$$ is increasing on the stated interval it is enough to show that its derivative $$y'(\theta)\ge 0$$ for every $$\theta$$ in that interval.

Step 1: Differentiate the first term

Let $$f(\theta)=\dfrac{4\sin\theta}{2+\cos\theta}.$$ Using the quotient rule,

$$ \begin{aligned} f'(\theta)&=\dfrac{(4\cos\theta)(2+\cos\theta)-4\sin\theta(-\sin\theta)}{(2+\cos\theta)^2}\\ &=\dfrac{4\cos\theta(2+\cos\theta)+4\sin^2\theta}{(2+\cos\theta)^2}.\end{aligned}$$

Replace $$\sin^2\theta$$ by $$1-\cos^2\theta$$:

$$ \begin{aligned} f'(\theta)&=\dfrac{4\bigl[\cos\theta(2+\cos\theta)+(1-\cos^2\theta)\bigr]}{(2+\cos\theta)^2}\\[2pt] &=\dfrac{4\bigl(2\cos\theta+1\bigr)}{(2+\cos\theta)^2}\\[2pt] &=\dfrac{8\cos\theta+4}{(2+\cos\theta)^2}. \end{aligned}$$

Step 2: Differentiate the whole function

Because $$y(\theta)=f(\theta)-\theta$$, we have

$$y'(\theta)=f'(\theta)-1=\frac{8\cos\theta+4}{(2+\cos\theta)^2}-1.$$

Put everything over the common denominator $$(2+\cos\theta)^2$$:

$$ \begin{aligned} y'(\theta)&=\frac{8\cos\theta+4-(2+\cos\theta)^2}{(2+\cos\theta)^2}\\[2pt] &=\frac{8\cos\theta+4-\bigl(4+4\cos\theta+\cos^2\theta\bigr)}{(2+\cos\theta)^2}\\[2pt] &=\frac{4\cos\theta-\cos^2\theta}{(2+\cos\theta)^2}\\[2pt] &=\frac{\cos\theta\,(4-\cos\theta)}{(2+\cos\theta)^2}. \end{aligned}$$

Step 3: Sign of the derivative on $$\left[0,\dfrac{\pi}{2}\right]$$

• For $$\theta\in\left[0,\dfrac{\pi}{2}\right]$$ we have $$\cos\theta\ge 0.$$
• Also $$4-\cos\theta\ge 4-1=3>0.$$
• The denominator $$(2+\cos\theta)^2>0$$ everywhere.

Therefore $$y'(\theta)=\dfrac{\cos\theta\,(4-\cos\theta)}{(2+\cos\theta)^2}\ge 0$$ for every $$\theta\in\left[0,\dfrac{\pi}{2}\right]$$ and is zero only at $$\theta=\dfrac{\pi}{2}$$.

Conclusion
Since $$y'(\theta)\ge 0$$ throughout the given interval, the function $$y=\dfrac{4\sin\theta}{2+\cos\theta}-\theta$$ is increasing (more precisely, non-decreasing) on $$\left[0,\dfrac{\pi}{2}\right].$$

Objective : Show that the logarithmic function is increasing on the interval $$(0,\infty)$$.

Let $$f:(0,\infty)\to\mathbb R$$ be given by $$f(x)=\log_b x$$, where the base satisfies $$b>1$$. (The natural logarithm $$\ln x$$ is the particular case $$b=e$$.)

Step 1 : Differentiate.

Using the differentiation rule for the logarithm,

$$f'(x)=\dfrac{1}{x\ln b}.$$

Step 2 : Examine the sign of the derivative.

• For every $$x\in(0,\infty)$$ we have $$x>0$$, so $$1/x>0$$.
• The constant $$\ln b$$ is positive because $$b>1$$.

Therefore the product $$1/(x\ln b)$$ is positive for every $$x>0$$; that is,

$$f'(x)>0\quad\text{for all }x\in(0,\infty).$$

Step 3 : Invoke the Monotonicity Test.

For a differentiable function, if $$f'(x)>0$$ throughout an interval, the function is strictly increasing on that interval. Since $$f'(x)>0$$ on $$(0,\infty)$$, the function $$f(x)=\log_b x$$ is strictly increasing on $$(0,\infty)$$.

Conclusion : The logarithmic function (for any base $$b>1$$, including the natural logarithm) increases as its argument moves from left to right over the entire positive real line.

Objective Prove that the function $$f(x)=x^2-x+1$$ is neither strictly increasing nor strictly decreasing on the open interval $$(-1,1)$$.

Step 1  — Recall the definitions
A function $$f$$ is
  • strictly increasing on an interval if for every pair $$x_1,x_2$$ in that interval, $$x_1<x_2\;\Rightarrow\;f(x_1)<f(x_2)$$.
  • strictly decreasing on an interval if for every pair $$x_1,x_2$$ in that interval, $$x_1<x_2\;\Rightarrow\;f(x_1)>f(x_2)$$.
A convenient test uses the first derivative: if $$f'(x)>0$$ (resp. $$<0$$) at every point of an interval, then $$f$$ is strictly increasing (resp. strictly decreasing) there.

Step 2  — Compute the derivative of $$f$$
$$f(x)=x^2-x+1\;\Longrightarrow\;f'(x)=\frac{d}{dx}(x^2)-\frac{d}{dx}(x)+\frac{d}{dx}(1)=2x-1.$$

Step 3  — Locate the critical point(s)
A critical point occurs where $$f'(x)=0$$ or $$f'$$ is undefined. Here $$f'(x)=2x-1$$ is defined for every real $$x$$, so we only have to solve

$$2x-1=0\;\Longrightarrow\;x=\frac12.$$

This point $$\tfrac12$$ lies inside the given interval $$(-1,1)$$.

Step 4  — Sign analysis of $$f'(x)$$ on $$(-1,1)$$

Thus

• $$f'(x)<0$$ for every $$x\in(-1,\tfrac12)$$  →  $$f$$ is strictly decreasing on $$(-1,\tfrac12)$$.
• $$f'(x)>0$$ for every $$x\in(\tfrac12,1)$$  →  $$f$$ is strictly increasing on $$(\tfrac12,1)$$.

Step 5  — Combine the results
Because $$f'(x)$$ changes from negative to positive at $$x=\tfrac12$$, the function first falls and then rises. Therefore there is no single monotonic behaviour throughout the whole interval $$(-1,1)$$.

Step 6  — Explicit counter-example (optional check)
Choose three points in $$(-1,1)$$, say $$x_1=-0.5, x_2=0.2, x_3=0.8$$. Compute:

$$f(-0.5)=(-0.5)^2-(-0.5)+1=0.25+0.5+1=1.75,\qquad f(0.2)=(0.2)^2-0.2+1=0.04-0.2+1=0.84,\qquad f(0.8)=(0.8)^2-0.8+1=0.64-0.8+1=0.84.$$

We have $$f(-0.5)>f(0.2)$$ but $$f(0.2)=f(0.8)$$, refuting both strict increase and strict decrease on the full interval.

Conclusion
The function $$f(x)=x^2-x+1$$ decreases on $$(-1,\tfrac12)$$ and increases on $$(\tfrac12,1)$$, so it is neither strictly increasing nor strictly decreasing on the entire interval $$(-1,1)$$.

We want the quadratic $$f(x) = x^2 + ax + 1$$ to be increasing on the closed interval $$[1, 2]$$.

Step 1: First derivative

$$f'(x) = 2x + a$$

Step 2: Condition for increasing behaviour

A differentiable function is increasing on an interval when $$f'(x) \ge 0$$ at every point of that interval. So we require

$$2x + a \ge 0 \quad \text{for all } x \in [1, 2].$$

Step 3: Apply the condition

The derivative $$f'(x) = 2x + a$$ is itself an increasing function of $$x$$ (its slope is $$2 > 0$$), so on $$[1, 2]$$ it takes its smallest value at the left end point $$x = 1$$. It is therefore enough to require

$$f'(1) \ge 0 \quad\Rightarrow\quad 2(1) + a \ge 0 \quad\Rightarrow\quad a \ge -2.$$

Step 4: Verification

If $$a \ge -2$$, then for every $$x \in [1, 2]$$ we have $$f'(x) = 2x + a \ge 2(1) + a \ge 0$$, so $$f$$ is increasing on $$[1, 2]$$. If instead $$a$$ were less than $$-2$$, then $$f'(1) = 2 + a$$ would be negative and $$f$$ would not be increasing near $$x = 1$$.

Therefore $$f(x) = x^2 + ax + 1$$ is increasing on $$[1, 2]$$ precisely for all real numbers $$a \ge -2$$.

Given: $$f(x)=x+\frac{1}{x}$$ with domain $$\mathbb{R}\setminus\{0\}$$.

Let $$I$$ be any interval such that $$I\cap[-1,1]=\varnothing$$. Then necessarily $$I\subset(-\infty,-1)$$ or $$I\subset(1,\infty)$$ (because an interval that does not meet $$[-1,1]$$ lies entirely to its left or to its right).

Step 1 – Derivative of $$f$$

For every $$x\neq0$$,

$$f'(x)=\dfrac{d}{dx}\Bigl(x+\frac{1}{x}\Bigr)=1-\frac{1}{x^{2}}=\frac{x^{2}-1}{x^{2}}.$$

Step 2 – Sign of the derivative on $$I$$

• When $$x<-1$$, we have $$x^{2}>1$$ and $$x^{2}>0$$, hence $$x^{2}-1>0$$ and so $$f'(x)=\dfrac{x^{2}-1}{x^{2}}>0$$.
• When $$x>1$$, again $$x^{2}>1$$ and $$x^{2}>0$$; therefore $$f'(x)=\dfrac{x^{2}-1}{x^{2}}>0$$.

Thus $$f'(x)>0$$ for every $$x$$ in $$(-\infty,-1)\cup(1,\infty)$$, and in particular for every $$x\in I$$.

Step 3 – Conclusion

Because the derivative is positive at every point of $$I$$, the function $$f$$ is strictly increasing on the interval $$I$$.

Hence, for any interval disjoint from $$[-1,1]$$, the function $$x+\dfrac1x$$ is increasing.

Let $$f(x) = \log \sin x$$. The argument of the logarithm must be positive, and since $$\sin x \gt 0$$ for every $$x \in (0, \pi)$$, the function is defined on the whole interval $$(0, \pi)$$, which contains both intervals in the question.

Step 1 — Differentiate.

Using the chain rule with $$\dfrac{d}{dx}(\log u) = \dfrac{1}{u} \cdot \dfrac{du}{dx}$$ and $$u = \sin x$$,

$$f'(x) = \dfrac{1}{\sin x} \cdot \cos x = \cot x.$$

Step 2 — Sign of $$f'(x)$$ on $$\left(0, \dfrac{\pi}{2}\right)$$.

Here the angle lies in the first quadrant, so

$$\sin x \gt 0 \quad \text{and} \quad \cos x \gt 0.$$

Therefore

$$f'(x) = \cot x = \dfrac{\cos x}{\sin x} \gt 0 \quad \text{for every } x \in \left(0, \dfrac{\pi}{2}\right).$$

Step 3 — Sign of $$f'(x)$$ on $$\left(\dfrac{\pi}{2}, \pi\right)$$.

Here the angle lies in the second quadrant, so

$$\sin x \gt 0 \quad \text{but} \quad \cos x \lt 0.$$

Therefore

$$f'(x) = \cot x = \dfrac{\cos x}{\sin x} \lt 0 \quad \text{for every } x \in \left(\dfrac{\pi}{2}, \pi\right).$$

Step 4 — Conclusion.

By the monotonicity test, if $$f'(x) \gt 0$$ throughout an interval then $$f$$ is strictly increasing there; if $$f'(x) \lt 0$$ throughout an interval then $$f$$ is strictly decreasing there.

Since $$f'(x) \gt 0$$ on $$\left(0, \dfrac{\pi}{2}\right)$$, the function $$f(x) = \log \sin x$$ is increasing on $$\left(0, \dfrac{\pi}{2}\right)$$.

Since $$f'(x) \lt 0$$ on $$\left(\dfrac{\pi}{2}, \pi\right)$$, the function $$f(x) = \log \sin x$$ is decreasing on $$\left(\dfrac{\pi}{2}, \pi\right)$$. Hence proved.

Let $$f(x) = \log |\cos x|$$. The function is defined wherever $$\cos x \ne 0$$. Both intervals in the question, $$\left(0, \dfrac{\pi}{2}\right)$$ and $$\left(\dfrac{3\pi}{2}, 2\pi\right)$$, lie inside the domain.

Step 1 — Remove the modulus.

On $$\left(0, \dfrac{\pi}{2}\right)$$ the angle lies in the first quadrant, so $$\cos x \gt 0$$. On $$\left(\dfrac{3\pi}{2}, 2\pi\right)$$ the angle lies in the fourth quadrant, so again $$\cos x \gt 0$$. Hence on both intervals $$|\cos x| = \cos x$$, and

$$f(x) = \log(\cos x).$$

Step 2 — Differentiate.

By the chain rule,

$$f'(x) = \dfrac{1}{\cos x} \cdot (-\sin x) = -\tan x.$$

Step 3 — Sign of $$f'(x)$$ on $$\left(0, \dfrac{\pi}{2}\right)$$.

For $$x \in \left(0, \dfrac{\pi}{2}\right)$$, the angle lies in the first quadrant, where $$\tan x \gt 0$$. Therefore

$$f'(x) = -\tan x \lt 0 \quad \text{for every } x \in \left(0, \dfrac{\pi}{2}\right),$$

so $$f$$ is strictly decreasing on $$\left(0, \dfrac{\pi}{2}\right)$$.

Step 4 — Sign of $$f'(x)$$ on $$\left(\dfrac{3\pi}{2}, 2\pi\right)$$.

For $$x \in \left(\dfrac{3\pi}{2}, 2\pi\right)$$, the angle lies in the fourth quadrant, where $$\sin x \lt 0$$ and $$\cos x \gt 0$$, so $$\tan x = \dfrac{\sin x}{\cos x} \lt 0$$. Therefore

$$f'(x) = -\tan x \gt 0 \quad \text{for every } x \in \left(\dfrac{3\pi}{2}, 2\pi\right),$$

so $$f$$ is strictly increasing on $$\left(\dfrac{3\pi}{2}, 2\pi\right)$$.

Conclusion. $$f(x) = \log |\cos x|$$ is decreasing on $$\left(0, \dfrac{\pi}{2}\right)$$ and increasing on $$\left(\dfrac{3\pi}{2}, 2\pi\right)$$. Hence proved.

Given function : $$f(x)=x^3-3x^2+3x-100$$, defined for every real number.

To prove that $$f(x)$$ is increasing on the entire real line $$\mathbf R$$ we use the first–derivative test for monotonicity :

• If $$f'(x)\ge 0$$ for all $$x\in\mathbf R$$, then $$f(x)$$ is non-decreasing (increasing) on $$\mathbf R$$.
• If, in addition, $$f'(x)=0$$ only at isolated points, the function is in fact strictly increasing.

Step 1 – Differentiate

Differentiate term–by–term:

$$f'(x)=\dfrac{d}{dx}\,(x^3)-\dfrac{d}{dx}\,(3x^2)+\dfrac{d}{dx}\,(3x)-\dfrac{d}{dx}\,(100).$$

Compute each derivative:

• $$\dfrac{d}{dx}\,(x^3)=3x^2$$ (power rule),
• $$\dfrac{d}{dx}\,(3x^2)=3\cdot 2x=6x$$,
• $$\dfrac{d}{dx}\,(3x)=3$$,
• $$\dfrac{d}{dx}\,(100)=0$$ (constant).

Hence

$$f'(x)=3x^2-6x+3.$$

Step 2 – Factorise $$f'(x)$$ and test its sign

First factor out the common factor $$3$$:

$$f'(x)=3\,(x^2-2x+1).$$

Recognise the quadratic inside the brackets as a perfect square:

$$x^2-2x+1=(x-1)^2.$$ Therefore,

$$f'(x)=3\,(x-1)^2.$$

For any real number $$x$$, the square $$(x-1)^2$$ is always non-negative, i.e. $$ (x-1)^2\ge 0.$$ Multiplying by the positive constant $$3$$ does not change the sign, so

$$f'(x)=3\,(x-1)^2\ge 0\quad\text{for every }x\in\mathbf R.$$

Moreover, $$f'(x)=0$$ only when $$(x-1)^2=0\implies x=1.$$ Thus the derivative is zero at exactly one point and positive everywhere else.

Step 3 – Conclude monotonicity

Since $$f'(x)\ge 0$$ for all $$x$$, by the derivative test $$f(x)$$ is increasing (non-decreasing) on $$\mathbf R$$.

Because $$f'(x)>0$$ except at the single point $$x=1$$, the function never decreases; any two points with $$x_2>x_1$$ give

$$f(x_2)-f(x_1)=f'(c)(x_2-x_1)\ge 0\quad(\text{MVT with }c\in(x_1,x_2)).$$

Hence $$f(x)$$ is, in fact, strictly increasing on the whole real line.

Final Statement : The function $$f(x)=x^3-3x^2+3x-100$$ is increasing on $$\mathbf R.$$

Given function: $$f(x)=x^2$$, where $$x\in\mathbf{R}$$.

We have to find its extreme values (maximum or minimum).

1. Find the first derivative
   $$\dfrac{df}{dx}=f'(x)=2x$$.
2. Find critical (stationary) points
   Set $$f'(x)=0$$:
   $$2x=0\; \Longrightarrow\; x=0$$.
   Thus, the only critical point in $$\mathbf{R}$$ is $$x=0$$.
3. Apply the Second Derivative Test
   Compute $$f''(x)=\dfrac{d}{dx}(2x)=2$$, which is a positive constant.
   At $$x=0$$, $$f''(0)=2>0$$  ⇒  $$f$$ has a point of local minimum at $$x=0$$.
4. Value of the function at the stationary point
   $$f(0)=0^2=0$$.
   Hence the minimum value of $$f$$ is $$0$$, attained at $$x=0$$.
5. Existence of a maximum value
   Because the domain is the entire real line, we can make $$x^2$$ as large as we like by choosing $$|x|$$ sufficiently large.
   Formally, for every real number $$M$$ there exists $$x$$ (take $$x=\sqrt{M}$$ or $$x=-\sqrt{M}$$) such that $$f(x)=x^2 > M$$.
   Therefore, $$f$$ is unbounded above and has no maximum value.

Conclusion: $$f(x)=x^2$$ attains a minimum value $$0$$ at $$x=0$$ and has no maximum value on $$\mathbf{R}$$.

The function is $$f(x) = |x|$$ with domain $$\mathbf{R}$$.

Step 1 — Piecewise form.

$$f(x) = \begin{cases} -x, & \text{if } x \lt 0, \\ x, & \text{if } x \ge 0. \end{cases}$$

Step 2 — Derivative on each branch.

• For $$x \lt 0$$, $$f(x) = -x$$, so $$f'(x) = -1$$.
• For $$x \gt 0$$, $$f(x) = x$$, so $$f'(x) = 1$$.
• At $$x = 0$$, the left-hand derivative is $$-1$$ and the right-hand derivative is $$+1$$, so $$f'(0)$$ does not exist. The function is, however, continuous at $$x = 0$$.

Since $$f'(x) \ne 0$$ for every $$x \ne 0$$, the only critical point is $$x = 0$$.

Step 3 — Nature of the critical point.

As $$x$$ moves through $$0$$, the derivative changes from $$-1$$ (negative) to $$+1$$ (positive). That is, $$f$$ is decreasing on $$(-\infty, 0)$$ and increasing on $$(0, \infty)$$. By the first-derivative test, $$x = 0$$ is a point of local (and in fact global) minimum.

Step 4 — Minimum value.

$$f(0) = |0| = 0.$$ The minimum value of $$f$$ is $$0$$, attained at $$x = 0$$.

Step 5 — Maximum value.

As $$|x| \to \infty$$, $$f(x) = |x| \to \infty$$. So $$f$$ is unbounded above and has no maximum value on $$\mathbf{R}$$.

Conclusion. Minimum value $$= 0$$ at $$x = 0$$; no maximum value exists.

Step 1 : Check differentiability on the given interval

The function is $$f(x)=x$$ defined for every $$x\in(0,1)$$. It is a polynomial, hence continuous and differentiable on the open interval.

Step 2 : Find critical points

Compute the first derivative:

$$f'(x)=1$$ for all $$x\in(0,1)$$.

A critical point occurs where $$f'(x)=0$$ or $$f'$$ is undefined. Since $$f'(x)=1\neq0$$ everywhere and is defined everywhere on $$(0,1)$$, there are no critical points inside the interval.

Step 3 : Analyse monotonicity

Because $$f'(x)=1>0$$ for every $$x$$ in the domain, the function is strictly increasing on the whole interval $$(0,1)$$.

Step 4 : Examine the end–points

The closed–interval method would compare the values at critical points and at the end–points, but here the end–points $$x=0$$ and $$x=1$$ are not included in the domain.

• As $$x\to0^+$$, $$f(x)=x\to0$$.
• As $$x\to1^-$$, $$f(x)=x\to1$$.

However, the values $$0$$ and $$1$$ are not obtained because the corresponding $$x$$ values are not contained in $$(0,1)$$.

Step 5 : Conclusion

Since the strictly increasing function has no critical point and the end–point values are not attained, it possesses neither a maximum nor a minimum value on $$(0,1)$$ (though its supremum is $$1$$ and its infimum is $$0$$).

Given $$f(x) = x^3 - 3x + 3$$.

Step 1 — Critical points.

Differentiate:

$$f'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x - 1)(x + 1).$$

Set $$f'(x) = 0$$:

$$3(x - 1)(x + 1) = 0 \quad\Rightarrow\quad x = 1 \ \text{or}\ x = -1.$$

Thus the two critical points are $$x = -1$$ and $$x = 1$$.

Step 2 — Second-derivative test.

$$f''(x) = \dfrac{d}{dx}(3x^2 - 3) = 6x.$$

• At $$x = 1$$: $$f''(1) = 6(1) = 6 > 0$$, so $$x = 1$$ is a point of local minimum.
• At $$x = -1$$: $$f''(-1) = 6(-1) = -6 < 0$$, so $$x = -1$$ is a point of local maximum.

Step 3 — Function values at the critical points.

$$f(1) = 1^3 - 3(1) + 3 = 1 - 3 + 3 = 1,$$

$$f(-1) = (-1)^3 - 3(-1) + 3 = -1 + 3 + 3 = 5.$$

Conclusion. $$f$$ has a local maximum value $$5$$ at $$x = -1$$ and a local minimum value $$1$$ at $$x = 1$$.

Given function
$$f(x)=2x^3-6x^2+6x+5$$

1 ‣ First derivative & critical points

$$f'(x)=\frac{d}{dx}\,[2x^3-6x^2+6x+5]=6x^2-12x+6$$ $$f'(x)=6(x^2-2x+1)=6(x-1)^2$$

For stationary (critical) points set $$f'(x)=0$$:

$$6(x-1)^2=0\;\Longrightarrow\;x=1$$

Thus the only stationary point is at $$x=1$$. Its ordinate is

$$f(1)=2(1)^3-6(1)^2+6(1)+5=2-6+6+5=7$$

2 ‣ Second derivative test

$$f''(x)=\frac{d}{dx}\,f'(x)=\frac{d}{dx}\,[6(x-1)^2]=12(x-1)$$ $$f''(1)=12(1-1)=0$$

Because $$f''(1)=0$$, the usual second-derivative test is inconclusive; we must examine further.

3 ‣ Sign of f′(x) on either side of x = 1

Thus $$f'(x)\gt0$$ on both sides of $$x=1$$ (it merely touches zero at the point).
The function is increasing before and after $$x=1$$, so the point $$x=1$$ is neither a local maximum nor a local minimum.

4 ‣ Higher-order confirmation

The third derivative is

$$f'''(x)=\frac{d}{dx}\,f''(x)=12\neq0$$

The first non-zero derivative after $$f'(x)$$ is of odd order (3^rd) and positive; therefore $$x=1$$ is a point of inflexion with horizontal tangent, not an extremum.

Conclusion

The function $$f(x)=2x^3-6x^2+6x+5$$ has no points of local maximum or local minimum. The stationary point $$(1,7)$$ is a point of inflexion.

Given $$f(x) = 3 + |x|$$ on $$\mathbf{R}$$.

Step 1 — Piecewise form.

$$f(x) = \begin{cases} 3 + x, & \text{if } x \ge 0, \\ 3 - x, & \text{if } x \lt 0. \end{cases}$$

Step 2 — Derivative on each branch.

• For $$x \gt 0$$, $$f'(x) = \dfrac{d}{dx}(3 + x) = 1$$.
• For $$x \lt 0$$, $$f'(x) = \dfrac{d}{dx}(3 - x) = -1$$.
• At $$x = 0$$ the left-hand derivative is $$-1$$ and the right-hand derivative is $$+1$$, so $$f'(0)$$ does not exist. The function is continuous at $$x = 0$$.

Step 3 — Critical points.

$$f'(x) \ne 0$$ for any $$x \ne 0$$, and $$f'(0)$$ does not exist. So the only critical point is $$x = 0$$.

Step 4 — First-derivative test at $$x = 0$$.

• To the left of $$0$$: $$f'(x) = -1 \lt 0$$, so $$f$$ is decreasing.
• To the right of $$0$$: $$f'(x) = +1 \gt 0$$, so $$f$$ is increasing.

The derivative changes sign from negative to positive at $$x = 0$$, so by the first-derivative test, $$x = 0$$ is a point of local minimum.

Step 5 — Minimum value.

$$f(0) = 3 + |0| = 3.$$

Therefore, the function attains a local minimum value $$3$$ at $$x = 0$$.

Given $$f(x) = 3x^4 + 4x^3 - 12x^2 + 12$$.

Step 1 — First derivative.

$$f'(x) = 12x^3 + 12x^2 - 24x.$$

Step 2 — Critical points.

Set $$f'(x) = 0$$:

$$12x^3 + 12x^2 - 24x = 12x(x^2 + x - 2) = 12x(x + 2)(x - 1) = 0.$$

Therefore the critical points are $$x = -2$$, $$x = 0$$, and $$x = 1$$.

Step 3 — Second-derivative test.

$$f''(x) = 36x^2 + 24x - 24.$$

• At $$x = 0$$: $$f''(0) = 36(0)^2 + 24(0) - 24 = -24 < 0$$, so $$x = 0$$ is a point of local maximum.
• At $$x = -2$$: $$f''(-2) = 36(4) + 24(-2) - 24 = 144 - 48 - 24 = 72 > 0$$, so $$x = -2$$ is a point of local minimum.
• At $$x = 1$$: $$f''(1) = 36(1) + 24(1) - 24 = 36 > 0$$, so $$x = 1$$ is a point of local minimum.

Step 4 — Function values at the critical points.

• $$f(0) = 3(0)^4 + 4(0)^3 - 12(0)^2 + 12 = 12.$$
• $$f(-2) = 3(16) + 4(-8) - 12(4) + 12 = 48 - 32 - 48 + 12 = -20.$$
• $$f(1) = 3(1) + 4(1) - 12(1) + 12 = 3 + 4 - 12 + 12 = 7.$$

Conclusion. $$f$$ has a local maximum value $$12$$ at $$x = 0$$, and local minimum values $$-20$$ at $$x = -2$$ and $$7$$ at $$x = 1$$.

Given function
$$f(x)=2x^3-6x^2+6x+5$$

Step 1 : Find the first derivative
$$f'(x)=\dfrac{\mathrm d}{\mathrm dx}\,(2x^3-6x^2+6x+5)=6x^2-12x+6$$

Step 2 : Locate critical points
Set $$f'(x)=0$$ :
$$6x^2-12x+6=0\;\Longrightarrow\;x^2-2x+1=0\;(\text{divide by }6)$$ $$x^2-2x+1=(x-1)^2=0\;\Longrightarrow\;x=1$$ Thus the only stationary point is $$x=1$$.

Step 3 : Second derivative test
$$f''(x)=\dfrac{\mathrm d}{\mathrm dx}\,(6x^2-12x+6)=12x-12$$
At $$x=1$$ : $$f''(1)=12(1)-12=0$$
Since the second derivative is zero, this test is inconclusive.

Step 4 : First derivative sign analysis
Rewrite $$f'(x)$$ : $$f'(x)=6(x-1)^2\ge 0\;\text{for all }x$$.
• For $$x\ne 1$$, $$f'(x)>0$$ (positive).
• At $$x=1$$, $$f'(1)=0$$.

Hence the derivative does not change sign as $$x$$ passes through $$1$$: it is positive on both sides and merely touches zero at the point. Therefore the function is increasing throughout, with a short horizontal tangent at $$x=1$$.

Step 5 : Conclusion
Because $$f'(x)$$ does not change from positive to negative (or vice-versa), the point $$x=1$$ is neither a local maximum nor a local minimum. It is, in fact, a stationary point of inflection.

Value of the function at the stationary point
$$f(1)=2(1)^3-6(1)^2+6(1)+5=7$$

Final result
The function $$f(x)=2x^3-6x^2+6x+5$$ has no points of local maxima or local minima.

Let the required positive numbers be $$x$$ and $$y$$.

According to the given condition, their sum is fixed:

$$x + y = 15 \Rightarrow y = 15 - x, \; 0 < x < 15.$$

We want to minimise the sum of their squares

$$S(x) = x^2 + y^2 = x^2 + (15 - x)^2.$$

Simplify the objective function:

$$S(x) = x^2 + (15 - x)^2 = x^2 + 225 - 30x + x^2 = 2x^2 - 30x + 225.$$

To find the critical point, differentiate $$S(x)$$ with respect to $$x$$:

$$\frac{\mathrm dS}{\mathrm dx} = 4x - 30.$$

Set the derivative to zero for an extremum:

$$4x - 30 = 0 \;\Rightarrow\; x = \frac{30}{4} = \frac{15}{2} = 7.5.$$

Compute the second derivative to confirm the nature of the extremum:

$$\frac{\mathrm d^2 S}{\mathrm dx^2} = 4 > 0,$$

which shows that $$S(x)$$ attains a minimum at $$x = 7.5$$.

Using $$y = 15 - x$$, we get

$$y = 15 - 7.5 = 7.5.$$

Hence, the two positive numbers that minimise the sum of squares are

$$7.5 \text{ and } 7.5.$$

(The minimum value of the sum of squares is $$2\times 7.5^2 = 112.5$$, although the problem only asks for the numbers.)

Step 1 : Write the distance in terms of a single variable
Let $$P(0,c)$$ be the given point with $$\tfrac12\le c\le 5$$ and $$Q(x,x^2)$$ be any point on the parabola $$y=x^2$$. The distance $$PQ$$ is

$$D(x)=\sqrt{(x-0)^2+(x^2-c)^2}=\sqrt{x^2+(x^2-c)^2}.$$

Because the square-root function is increasing, minimising $$D(x)$$ is equivalent to minimising the squared distance

$$f(x)=x^2+(x^2-c)^2.$$

Step 2 : Simplify the function to be minimised

$$f(x)=x^2+(x^4-2cx^2+c^2)=x^4+(1-2c)x^2+c^2.$$

Step 3 : Find the critical points

Differentiate:

$$f'(x)=4x^3+2(1-2c)x=2x\bigl(2x^2+1-2c\bigr).$$

Set $$f'(x)=0$$:

• $$x=0,$$ or
• $$2x^2+1-2c=0\;\Longrightarrow\;x^2=c-\tfrac12\;(\text{possible because }c\ge\tfrac12).$$

Hence the critical points are
$$x=0,\;x=\pm\sqrt{c-\tfrac12}. $$

Step 4 : Classify the critical points

Second derivative:

$$f''(x)=12x^2+2(1-2c).$$

• At $$x=\pm\sqrt{c-\tfrac12}$$:
  $$f''=12(c-\tfrac12)+2(1-2c)=8c-4=4(2c-1)\ge0,$$ so these points give minima.
• At $$x=0$$:
  $$f''(0)=2(1-2c)\le0,$$ which is negative for $$c>\tfrac12$$ (zero at $$c=\tfrac12$$). Thus $$x=0$$ is a maximum for $$c>\tfrac12$$ and only a point of inflection when $$c=\tfrac12$$.

Step 5 : Compute the minimum value of $$f$$

Put $$x^2=c-\tfrac12$$ in $$f(x)$$:

$$\begin{aligned} f_{\min}&=(c-\tfrac12)^2+(1-2c)(c-\tfrac12)+c^2\\ &=\bigl(c^2-c+\tfrac14\bigr)+\bigl(2c-2c^2-\tfrac12\bigr)+c^2\\ &= -c^2+c-\tfrac14+c^2 = c-\tfrac14. \end{aligned}$$

The minimum squared distance is therefore $$c-\tfrac14$$, and the minimum distance itself is

$$D_{\min}=\sqrt{c-\tfrac14}.$$ (The expression is non-negative because $$c\ge\tfrac12$$.)

Step 6 : Coordinates of the nearest point(s) on the parabola

At the minimum we have $$x=\pm\sqrt{c-\tfrac12}$$ and hence
$$y=x^2=c-\tfrac12.$$ Thus the two nearest points are
$$Q_1\bigl(\sqrt{c-\tfrac12},\,c-\tfrac12\bigr),\quad Q_2\bigl(-\sqrt{c-\tfrac12},\,c-\tfrac12\bigr).$$

Conclusion: For every $$\tfrac12\le c\le5,$$ the shortest distance from $$P(0,c)$$ to the parabola $$y=x^2$$ equals $$\sqrt{c-\tfrac14}$$.

Step 1 : Set–up a coordinate system
Let the ground be the X-axis. Place A at the origin, so $$A\,(0,0)$$. Since AB = 20 m, take $$B\,(20,0)$$.
Because the poles are vertical, their tops are
  $$P\,(0,16)$$ (AP = 16 m) and $$Q\,(20,22)$$ (BQ = 22 m).

Step 2 : Express RP² and RQ²
Let R be any point on AB with $$R\,(x,0)$$, where $$0 \le x \le 20$$.
Using the distance formula,

$$RP^2 = (x-0)^2 + (0-16)^2 = x^2 + 256,$$

$$RQ^2 = (x-20)^2 + (0-22)^2 = (x-20)^2 + 484.$$

Step 3 : Form the function to be minimised
Define $$f(x) = RP^2 + RQ^2$$:

$$f(x) = x^2 + 256 + (x-20)^2 + 484.$$

Simplify:

$$(x-20)^2 = x^2 - 40x + 400,$$

$$f(x) = x^2 + x^2 - 40x + 400 + 740 = 2x^2 - 40x + 1140.$$

Step 4 : Differentiate and find stationary point
$$f'(x) = \frac{d}{dx}(2x^2 - 40x + 1140) = 4x - 40.$$
Set $$f'(x)=0$$:

$$4x - 40 = 0 \;\Rightarrow\; x = 10.$$

Step 5 : Confirm it is a minimum
$$f''(x) = \frac{d}{dx}(4x - 40) = 4,$$ which is positive for all x. Hence $$x = 10$$ gives the minimum value of $$f(x).$$

Step 6 : Interpret the result
The required distance $$AR = x = 10\,\text{m}$$. Thus R is the midpoint of AB.

Let the trapezium be $$ABCD$$ with $$AB \parallel CD$$, where the three sides other than the base $$AB$$ each measure $$10\,\mathrm{cm}$$. Thus

• $$CD = 10\,\mathrm{cm}$$ (the second parallel side),
• $$AD = BC = 10\,\mathrm{cm}$$ (the non-parallel sides).

Because $$AD = BC$$, the trapezium is isosceles. Drop perpendiculars from $$C$$ and $$D$$ to the base $$AB$$; let their feet be $$M$$ and $$N$$ respectively, with $$N$$ on the side of $$A$$ and $$M$$ on the side of $$B$$. Then $$DN = CM = h$$ is the height.

By symmetry, $$AN = BM = s$$ for some $$s \ge 0$$, and the rectangle $$DNMC$$ has $$NM = CD = 10$$. Hence

$$AB = AN + NM + MB = s + 10 + s = 10 + 2s.$$

From the right triangle $$\triangle ADN$$ (with hypotenuse $$AD = 10$$, horizontal leg $$AN = s$$, vertical leg $$DN = h$$), the Pythagorean theorem gives

$$h = \sqrt{10^2 - s^2} = \sqrt{100 - s^2}, \quad 0 \le s \le 10.$$

The area of the trapezium is

$$A(s) = \tfrac{1}{2}\,(AB + CD)\,h = \tfrac{1}{2}\,(10 + 2s + 10)\,\sqrt{100 - s^2} = (10 + s)\sqrt{100 - s^2}.$$

Maximising $$A(s)$$.

Differentiate with respect to $$s$$:

$$\dfrac{dA}{ds} = \sqrt{100 - s^2} + (10 + s) \cdot \dfrac{-s}{\sqrt{100 - s^2}} = \dfrac{(100 - s^2) - s(10 + s)}{\sqrt{100 - s^2}} = \dfrac{100 - 10s - 2s^2}{\sqrt{100 - s^2}}.$$

Set $$\dfrac{dA}{ds} = 0$$:

$$100 - 10s - 2s^2 = 0 \quad\Longrightarrow\quad s^2 + 5s - 50 = 0.$$

By the quadratic formula,

$$s = \dfrac{-5 \pm \sqrt{25 + 200}}{2} = \dfrac{-5 \pm 15}{2}.$$

The meaningful (non-negative) root is $$s = 5\,\mathrm{cm}$$.

Confirm it is a maximum (first-derivative sign test).

The denominator $$\sqrt{100 - s^2} \gt 0$$ for $$0 \le s \lt 10$$. Hence the sign of $$\dfrac{dA}{ds}$$ is the sign of the numerator $$100 - 10s - 2s^2 = -2(s^2 + 5s - 50) = -2(s - 5)(s + 10)$$.

• For $$0 \lt s \lt 5$$: $$(s - 5) \lt 0$$ and $$(s + 10) \gt 0$$, so the numerator is positive. Hence $$\dfrac{dA}{ds} \gt 0$$ and $$A(s)$$ is increasing.
• For $$5 \lt s \lt 10$$: $$(s - 5) \gt 0$$ and $$(s + 10) \gt 0$$, so the numerator is negative. Hence $$\dfrac{dA}{ds} \lt 0$$ and $$A(s)$$ is decreasing.

Since $$\dfrac{dA}{ds}$$ changes from positive to negative at $$s = 5$$, $$A(s)$$ attains its maximum at $$s = 5$$.

Maximum dimensions:

• $$AB = 10 + 2(5) = 20\,\mathrm{cm}$$;
• $$h = \sqrt{100 - 25} = \sqrt{75} = 5\sqrt{3}\,\mathrm{cm}$$.

Maximum area:

$$A_{\max} = (10 + 5)\sqrt{75} = 15 \cdot 5\sqrt{3} = 75\sqrt{3}\,\mathrm{cm^2}.$$

Let the given right circular cone have base radius $$R$$ and height $$H$$. Inscribe a right circular cylinder with its axis along the cone's axis and its base on the cone's base. Let

• $$r$$ = radius of the inscribed cylinder,
• $$h$$ = height of the inscribed cylinder,

with $$0 \lt r \lt R$$.

Step 1 — Geometric relation between $$r$$ and $$h$$.

Consider the axial cross-section of the cone and cylinder. By the similarity of the two triangles cut by the cylinder's top face,

$$\dfrac{r}{R} = \dfrac{H - h}{H}.$$

Solving for $$h$$,

$$h = H\left(1 - \dfrac{r}{R}\right).$$

Step 2 — Curved surface area of the cylinder.

$$S = 2\pi r h = 2\pi r \cdot H\left(1 - \dfrac{r}{R}\right) = 2\pi H\left(r - \dfrac{r^2}{R}\right).$$

Step 3 — Maximise $$S$$ over $$0 \lt r \lt R$$.

Since $$2\pi H$$ is a positive constant, $$S$$ is maximised when

$$f(r) = r - \dfrac{r^2}{R}, \qquad 0 \lt r \lt R,$$

is maximised. Differentiate:

$$f'(r) = 1 - \dfrac{2r}{R}.$$

Set $$f'(r) = 0$$:

$$1 - \dfrac{2r}{R} = 0 \quad\Longrightarrow\quad r = \dfrac{R}{2}.$$

Step 4 — Confirm it is a maximum.

$$f''(r) = -\dfrac{2}{R} \lt 0 \quad \text{for every } r,$$

so $$r = \dfrac{R}{2}$$ is indeed a point of maximum of $$f$$ (and hence of $$S$$).

Conclusion. The right circular cylinder of greatest curved surface area inscribed in a given cone has radius $$r = \dfrac{R}{2}$$, i.e. half the radius of the cone. Hence proved.

Given function
$$f(x)=2x^{3}-15x^{2}+36x+1$$ is to be investigated for absolute (global) extrema on the closed interval $$[1,5]$$.

1. Find the critical points inside the interval

   Differentiate:
   $$f'(x)=\dfrac{d}{dx}\,(2x^{3}-15x^{2}+36x+1)=6x^{2}-30x+36$$

   Simplify the derivative:
   $$f'(x)=6x^{2}-30x+36=6\bigl(x^{2}-5x+6\bigr)=6(x-2)(x-3)$$

   Set $$f'(x)=0$$:
   $$6(x-2)(x-3)=0\;\Longrightarrow\;x=2,\;x=3$$

   Both critical points lie inside $$[1,5]$$, so they must be tested for absolute extrema.

2. Evaluate $$f(x)$$ at the endpoints and at the critical points

   Point	Computation of $$f(x)$$	Value
   $$x=1$$	$$2(1)^3-15(1)^2+36(1)+1$$	$$24$$
   $$x=2$$	$$2(2)^3-15(2)^2+36(2)+1$$	$$29$$
   $$x=3$$	$$2(3)^3-15(3)^2+36(3)+1$$	$$28$$
   $$x=5$$	$$2(5)^3-15(5)^2+36(5)+1$$	$$56$$

3. Select the largest and smallest values obtained

   • The greatest value is $$56$$ at $$x=5$$.
   • The least value is $$24$$ at $$x=1$$.

Hence, on the interval $$[1,5]$$ the function attains an absolute maximum of $$56$$ at $$x=5$$ and an absolute minimum of $$24$$ at $$x=1$$.

Function: $$f(x)=12x^{\frac{4}{3}}-6x^{\frac{1}{3}},\;x\in[-1,1].$$

1. First-order derivative

$$f'(x)=12\cdot\frac43x^{\frac13}-6\cdot\frac13x^{-\frac23}=16x^{\frac13}-2x^{-\frac23}.$$

2. Critical points in the open interval

(i) Solve $$f'(x)=0$$:

$$16x^{\frac13}-2x^{-\frac23}=0\;\Longrightarrow\;16x^{\frac13}=2x^{-\frac23}$$
$$8x^{\frac13}=x^{-\frac23}\;\Longrightarrow\;8x^{\frac13}\cdot x^{\frac23}=1$$
$$8x=1\;\Longrightarrow\;x=\tfrac18.$$

(ii) Derivative undefined at $$x=0$$ (because of $$x^{-\frac23}$$). The function itself is defined there, so $$x=0$$ is also a critical point.

Thus the candidates are $$x=-1,\;0,\;\tfrac18,\;1.$$

3. Function values at candidates

• $$f(-1)=12(-1)^{\frac43}-6(-1)^{\frac13}=12(1)-6(-1)=18.$$
• $$f(0)=0.$$
• $$f\!\left(\tfrac18\right)=12\left(\tfrac18\right)^{\frac43}-6\left(\tfrac18\right)^{\frac13}=12\cdot\tfrac1{16}-6\cdot\tfrac12=\tfrac34-3=-\tfrac94.$$
• $$f(1)=12-6=6.$$

4. Absolute extrema on $$[-1,1]$$

Maximum value $$18$$ occurs at $$x=-1.$$
Minimum value $$-\dfrac94$$ occurs at $$x=\dfrac18.$$

The helicopter is at a general point $$P(x, y)$$ on the curve $$y = x^2 + 7$$, and the soldier is at the fixed point $$S(3, 7)$$. We minimise the distance $$SP$$.

Step 1 — Express the distance in one variable.

$$SP^2 = (x - 3)^2 + (y - 7)^2.$$

Using $$y = x^2 + 7$$ gives $$y - 7 = x^2$$, so

$$SP^2 = (x - 3)^2 + (x^2)^2 = (x - 3)^2 + x^4.$$

Expanding,

$$D(x) := SP^2 = x^4 + x^2 - 6x + 9.$$

Since $$SP \ge 0$$, $$SP$$ is minimised exactly when $$D(x)$$ is minimised.

Step 2 — Critical points.

Differentiate:

$$D'(x) = 4x^3 + 2x - 6.$$

Set $$D'(x) = 0$$:

$$4x^3 + 2x - 6 = 0 \quad\Rightarrow\quad 2x^3 + x - 3 = 0.$$

Test $$x = 1$$: $$2(1)^3 + 1 - 3 = 0$$, so $$x = 1$$ is a root. Factor:

$$2x^3 + x - 3 = (x - 1)(2x^2 + 2x + 3).$$

The quadratic factor $$2x^2 + 2x + 3$$ has discriminant $$\Delta = 4 - 24 = -20 < 0$$, so it has no real zeros. The only real critical point is $$x = 1$$.

Step 3 — Verify it is a minimum.

$$D''(x) = 12x^2 + 2, \qquad D''(1) = 12 + 2 = 14 > 0,$$

so $$D$$ has a (global) minimum at $$x = 1$$.

Step 4 — Nearest point and minimum distance.

At $$x = 1$$, $$y = 1^2 + 7 = 8$$, so the nearest point on the curve is $$P(1, 8)$$.

$$SP = \sqrt{(1 - 3)^2 + (8 - 7)^2} = \sqrt{4 + 1} = \sqrt{5}.$$

Hence the least distance is $$\sqrt{5}$$ units, attained when the helicopter is at $$(1, 8)$$.

The profit earned from producing (or selling) $$x$$ units is $$p(x)=41-72x-18x^2$$.

Step 1 – First derivative
$$p'(x)=\frac{d}{dx}(41)-\frac{d}{dx}(72x)-\frac{d}{dx}(18x^2)=0-72-36x=-72-36x$$.

Step 2 – Stationary point
Set $$p'(x)=0$$:
$$-72-36x=0 \;\Longrightarrow\; -36x=72 \;\Longrightarrow\; x=-2$$.

Step 3 – Second derivative test
$$p''(x)=\frac{d}{dx}(-72-36x)=-36$$, which is negative for every $$x$$. Therefore, the stationary point at $$x=-2$$ gives a maximum profit.

Step 4 – Maximum profit
$$p(-2)=41-72(-2)-18(-2)^2=41+144-18\times4=41+144-72=113$$.

Hence, the greatest profit the company can earn is 113 (in the given currency units).

Let $$f(x) = 3x^4 - 8x^3 + 12x^2 - 48x + 25$$ on the closed interval $$[0, 3]$$.

Step 1 — Critical points.

$$f'(x) = 12x^3 - 24x^2 + 24x - 48 = 12(x^3 - 2x^2 + 2x - 4).$$

Let $$g(x) = x^3 - 2x^2 + 2x - 4$$. Setting $$g(x) = 0$$ and trying integer values, $$g(2) = 8 - 8 + 4 - 4 = 0$$, so $$x = 2$$ is a root. Dividing,

$$x^3 - 2x^2 + 2x - 4 = (x - 2)(x^2 + 2).$$

The factor $$x^2 + 2 > 0$$ for every real $$x$$, so its only real zero is $$x = 2$$. Hence the only critical point of $$f$$ in $$(0, 3)$$ is $$x = 2$$.

Step 2 — Function values at the critical point and the endpoints.

• $$f(0) = 25.$$
• $$f(2) = 3(16) - 8(8) + 12(4) - 48(2) + 25 = 48 - 64 + 48 - 96 + 25 = -39.$$
• $$f(3) = 3(81) - 8(27) + 12(9) - 48(3) + 25 = 243 - 216 + 108 - 144 + 25 = 16.$$

Step 3 — Absolute extrema on $$[0, 3]$$.

Comparing the three values, the absolute maximum value is $$25$$ at $$x = 0$$ and the absolute minimum value is $$-39$$ at $$x = 2$$.

Let $$f(x) = \sin 2x$$ for $$x \in [0, 2\pi]$$.

Step 1 — Stationary points.

$$f'(x) = 2\cos 2x.$$

Set $$f'(x) = 0$$: $$\cos 2x = 0$$, so $$2x = \dfrac{\pi}{2} + k\pi$$, i.e. $$x = \dfrac{\pi}{4} + \dfrac{k\pi}{2}$$ for some integer $$k$$.

The values lying in $$[0, 2\pi]$$ are

$$x = \dfrac{\pi}{4},\ \dfrac{3\pi}{4},\ \dfrac{5\pi}{4},\ \dfrac{7\pi}{4}.$$

Step 2 — Classify each stationary point using the second-derivative test.

$$f''(x) = -4\sin 2x.$$

• At $$x = \dfrac{\pi}{4}$$: $$2x = \dfrac{\pi}{2}$$, $$\sin 2x = 1$$, so $$f''\!\left(\dfrac{\pi}{4}\right) = -4(1) = -4 \lt 0$$. Hence $$x = \dfrac{\pi}{4}$$ is a local maximum, with value $$f\!\left(\dfrac{\pi}{4}\right) = 1$$.
• At $$x = \dfrac{3\pi}{4}$$: $$2x = \dfrac{3\pi}{2}$$, $$\sin 2x = -1$$, so $$f''\!\left(\dfrac{3\pi}{4}\right) = -4(-1) = 4 \gt 0$$. Hence $$x = \dfrac{3\pi}{4}$$ is a local minimum, with value $$f\!\left(\dfrac{3\pi}{4}\right) = -1$$.
• At $$x = \dfrac{5\pi}{4}$$: $$2x = \dfrac{5\pi}{2}$$, $$\sin 2x = \sin\dfrac{\pi}{2} = 1$$, so $$f''\!\left(\dfrac{5\pi}{4}\right) = -4(1) = -4 \lt 0$$. Hence $$x = \dfrac{5\pi}{4}$$ is a local maximum, with value $$f\!\left(\dfrac{5\pi}{4}\right) = 1$$.
• At $$x = \dfrac{7\pi}{4}$$: $$2x = \dfrac{7\pi}{2}$$, $$\sin 2x = \sin\dfrac{3\pi}{2} = -1$$, so $$f''\!\left(\dfrac{7\pi}{4}\right) = -4(-1) = 4 \gt 0$$. Hence $$x = \dfrac{7\pi}{4}$$ is a local minimum, with value $$f\!\left(\dfrac{7\pi}{4}\right) = -1$$.

Step 3 — Compare with the endpoint values.

$$f(0) = \sin 0 = 0, \qquad f(2\pi) = \sin 4\pi = 0.$$

The values at the local maxima are both equal to $$1$$, and since $$\sin 2x \le 1$$ for every $$x$$, this is the absolute maximum value of $$f$$ on $$[0, 2\pi]$$.

Conclusion. The maximum value of $$\sin 2x$$ on $$[0, 2\pi]$$ is $$1$$, attained at $$x = \dfrac{\pi}{4}$$ and $$x = \dfrac{5\pi}{4}$$.

Let $$f(x) = \sin x + \cos x$$ for all real $$x$$.

Step 1 — First derivative.

$$f'(x) = \cos x - \sin x.$$

Step 2 — Critical points.

Set $$f'(x) = 0$$:

$$\cos x - \sin x = 0 \quad\Longrightarrow\quad \tan x = 1 \quad\Longrightarrow\quad x = n\pi + \dfrac{\pi}{4}, \ n \in \mathbf{Z}.$$

Step 3 — Second-derivative test.

$$f''(x) = -\sin x - \cos x = -(\sin x + \cos x) = -f(x).$$

For even $$n$$ (i.e. $$x = 2k\pi + \dfrac{\pi}{4}$$):

$$f\!\left(2k\pi + \dfrac{\pi}{4}\right) = \sin\dfrac{\pi}{4} + \cos\dfrac{\pi}{4} = \dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2} = \sqrt{2},$$

so

$$f''\!\left(2k\pi + \dfrac{\pi}{4}\right) = -\sqrt{2} \lt 0.$$

By the second-derivative test, each such point is a local maximum.

For odd $$n$$ (i.e. $$x = (2k+1)\pi + \dfrac{\pi}{4} = 2k\pi + \dfrac{5\pi}{4}$$):

$$f\!\left(2k\pi + \dfrac{5\pi}{4}\right) = -\sqrt{2},$$

so

$$f''\!\left(2k\pi + \dfrac{5\pi}{4}\right) = -(-\sqrt{2}) = \sqrt{2} \gt 0.$$

By the second-derivative test, these points are local minima.

Step 4 — Maximum value.

The maximum value of $$f$$ is $$\sqrt{2}$$, attained at $$x = 2k\pi + \dfrac{\pi}{4}$$ for every integer $$k$$.

(Independent check: writing $$\sin x + \cos x = \sqrt{2}\sin\!\left(x + \dfrac{\pi}{4}\right)$$, the amplitude is $$\sqrt{2}$$, so $$f(x) \le \sqrt{2}$$, with equality attained.)

Let $$f(x)=2x^3-24x+107$$.

Step 1 : Compute the derivative.
$$f'(x)=\frac{d}{dx}\bigl(2x^3-24x+107\bigr)=6x^2-24=6(x^2-4)=6(x-2)(x+2).$$

Step 2 : Critical points.
Set $$f'(x)=0$$:

$$6(x-2)(x+2)=0\;\Rightarrow\;x=2\;\text{or}\;x=-2.$$

Part A : On the interval $$[1,3]$$

The only critical point that lies in $$[1,3]$$ is $$x=2$$. Evaluate $$f$$ at the two end-points and at the critical point:

• $$f(1)=2(1)^3-24(1)+107=2-24+107=85$$
• $$f(2)=2(8)-24(2)+107=16-48+107=75$$
• $$f(3)=2(27)-24(3)+107=54-72+107=89$$

Since $$89>85>75$$, the maximum value of $$f$$ on $$[1,3]$$ is $$89$$, attained at $$x=3$$.

Part B : On the interval $$[-3,-1]$$

The critical point lying in this interval is $$x=-2$$. Evaluate $$f$$ at the end-points and at $$x=-2$$:

• $$f(-3)=2(-27)-24(-3)+107=-54+72+107=125$$
• $$f(-2)=2(-8)-24(-2)+107=-16+48+107=139$$
• $$f(-1)=2(-1)-24(-1)+107=-2+24+107=129$$

Because $$139>129>125$$, the maximum value of $$f$$ on $$[-3,-1]$$ is $$139$$, attained at $$x=-2$$.

Conclusion.

• On $$[1,3]$$: maximum value $$=89$$ at $$x=3$$.
• On $$[-3,-1]$$: maximum value $$=139$$ at $$x=-2$$.

Let $$f(x) = x^4 - 62x^2 + ax + 9$$, and suppose $$f$$ attains its maximum on $$[0, 2]$$ at the interior point $$x = 1$$.

Step 1 — Stationary-point condition.

Since $$f$$ is differentiable on $$(0, 2)$$ and attains its maximum at the interior point $$x = 1$$, we must have $$f'(1) = 0$$.

$$f'(x) = 4x^3 - 124x + a.$$

Hence

$$f'(1) = 4(1)^3 - 124(1) + a = 4 - 124 + a = a - 120 = 0,$$

so $$a = 120$$.

Step 2 — Second-derivative test.

$$f''(x) = 12x^2 - 124.$$

At $$x = 1$$,

$$f''(1) = 12 - 124 = -112 < 0,$$

confirming that $$x = 1$$ is a point of local maximum.

Step 3 — Verification via endpoint comparison.

With $$a = 120$$,

• $$f(0) = 0 - 0 + 0 + 9 = 9$$;
• $$f(1) = 1 - 62 + 120 + 9 = 68$$;
• $$f(2) = 16 - 248 + 240 + 9 = 17$$.

The largest of these is $$f(1) = 68$$, so $$x = 1$$ is indeed the absolute maximum on $$[0, 2]$$.

Hence the required value of $$a$$ is $$120$$.

Let $$f(x) = x + \sin 2x$$ on $$[0, 2\pi]$$.

Step 1 — Stationary points.

$$f'(x) = 1 + 2\cos 2x.$$

Set $$f'(x) = 0$$: $$\cos 2x = -\dfrac{1}{2}$$. The general solution is $$2x = \pm\dfrac{2\pi}{3} + 2k\pi$$, i.e. $$x = \dfrac{\pi}{3} + k\pi$$ or $$x = \dfrac{2\pi}{3} + k\pi$$. The values lying in $$[0, 2\pi]$$ are

$$x = \dfrac{\pi}{3},\ \dfrac{2\pi}{3},\ \dfrac{4\pi}{3},\ \dfrac{5\pi}{3}.$$

Step 2 — Values of $$f$$ at the candidates and the endpoints.

Step 3 — Compare the values.

The largest of these is $$2\pi$$ (at $$x = 2\pi$$) and the smallest is $$0$$ (at $$x = 0$$).

Conclusion. Absolute maximum value $$= 2\pi$$ at $$x = 2\pi$$; absolute minimum value $$= 0$$ at $$x = 0$$.

Let the two numbers be $$x$$ and $$y$$, with $$x + y = 24$$.

We want to maximise their product

$$P = xy.$$

Using the constraint, $$y = 24 - x$$, so

$$P(x) = x(24 - x) = 24x - x^2.$$

Step 1 — First derivative.

$$\dfrac{dP}{dx} = 24 - 2x.$$

Set $$\dfrac{dP}{dx} = 0$$:

$$24 - 2x = 0 \quad\Longrightarrow\quad x = 12.$$

Step 2 — Second-derivative test.

$$\dfrac{d^2 P}{dx^2} = -2.$$

Since $$\dfrac{d^2 P}{dx^2} = -2 \lt 0$$ at every $$x$$, the function $$P(x)$$ attains a maximum at the stationary point $$x = 12$$.

Step 3 — Conclusion.

The other number is $$y = 24 - 12 = 12$$. Hence both numbers equal $$12$$, and the maximum product is

$$P_{\max} = 12 \times 12 = 144.$$

Let $$x$$ and $$y$$ be the required positive numbers with $$x + y = 60$$. We wish to maximise

$$P = xy^3.$$

From the constraint, $$x = 60 - y$$, so

$$P(y) = (60 - y)\,y^3 = 60y^3 - y^4.$$

Because both $$x$$ and $$y$$ are positive, the admissible domain is $$0 \lt y \lt 60$$. At the boundary points $$y = 0$$ and $$y = 60$$, $$P = 0$$, so any positive maximum must lie in the open interval.

Step 1 — First derivative.

$$P'(y) = 180y^2 - 4y^3 = 4y^2(45 - y).$$

Set $$P'(y) = 0$$:

$$4y^2(45 - y) = 0 \quad\Longrightarrow\quad y = 0 \ \text{or}\ y = 45.$$

Since $$y \gt 0$$, the admissible critical point is $$y = 45$$.

Step 2 — Second-derivative test.

$$P''(y) = 360y - 12y^2 = 12y(30 - y).$$

At $$y = 45$$,

$$P''(45) = 12(45)(30 - 45) = 540(-15) = -8100 \lt 0,$$

so $$P$$ attains a maximum at $$y = 45$$.

Step 3 — Corresponding value of $$x$$.

$$x = 60 - 45 = 15.$$

Therefore the two positive numbers that maximise $$xy^3$$ are $$x = 15$$ and $$y = 45$$.

Let $$x$$ and $$y$$ be positive numbers with $$x + y = 35$$. From the constraint, $$y = 35 - x$$, and we want to maximise

$$P(x) = x^2 (35 - x)^5, \qquad 0 \lt x \lt 35.$$

Step 1 — First derivative.

By the product rule,

$$P'(x) = 2x(35 - x)^5 + x^2 \cdot 5(35 - x)^4 \cdot (-1) = x(35 - x)^4\bigl[2(35 - x) - 5x\bigr].$$

Simplify the bracket: $$2(35 - x) - 5x = 70 - 7x = 7(10 - x)$$. Thus

$$P'(x) = 7x(35 - x)^4(10 - x).$$

Step 2 — Critical points.

Set $$P'(x) = 0$$. In the open interval $$0 \lt x \lt 35$$, both $$x \gt 0$$ and $$(35 - x)^4 \gt 0$$, so the only solution is

$$10 - x = 0 \quad\Longrightarrow\quad x = 10.$$

Step 3 — Confirm a maximum (first-derivative sign test).

• For $$0 \lt x \lt 10$$: $$(10 - x) \gt 0$$, so $$P'(x) \gt 0$$ — $$P$$ is increasing.
• For $$10 \lt x \lt 35$$: $$(10 - x) \lt 0$$, so $$P'(x) \lt 0$$ — $$P$$ is decreasing.

The derivative changes from positive to negative at $$x = 10$$, so $$P$$ attains its maximum there.

Step 4 — Corresponding $$y$$.

$$y = 35 - 10 = 25.$$

Therefore the required numbers are $$x = 10$$ and $$y = 25$$.

Let the required positive numbers be $$x$$ and $$(16-x)$$, so $$0<x<16$$.

Define the objective function

$$S(x)=x^{3}+(16-x)^{3}$$

Differentiate with respect to $$x$$:

$$\frac{dS}{dx}=3x^{2}-3(16-x)^{2}$$

For an extremum, set $$\frac{dS}{dx}=0$$:

$$3x^{2}-3(16-x)^{2}=0\;\Longrightarrow\;x^{2}=(16-x)^{2}$$

This implies either $$x=16-x$$ or $$x=-(16-x)$$. The second is impossible, hence

$$x=16-x\;\Longrightarrow\;2x=16\;\Longrightarrow\;x=8$$

The other number is $$16-8=8$$, so both numbers are positive.

Second-derivative test:

$$\frac{d^{2}S}{dx^{2}}=6x+6(16-x)=96>0$$

Since the second derivative is positive, $$x=8$$ gives a minimum.

Minimum value of the sum of cubes:

$$S_{\min}=8^{3}+8^{3}=2\times512=1024$$

Thus the two positive numbers whose sum is 16 and whose cubes have the minimum possible sum are $$8$$ and $$8$$.

Let a square of side $$18\,\text{cm}$$ be given. Cutting a square of side $$x\,\text{cm}$$ at each corner and folding up the flaps produces an open-top box whose:

• height is $$x$$,
• length and breadth are each $$(18-2x)$$ (because $$x$$ is removed from both ends).

Hence the volume of the box, as a function of $$x$$, is

$$V(x)=x(18-2x)^2.$$

Simplify to prepare for differentiation:

$$\begin{aligned} V(x)&=x(324-72x+4x^2)\\ &=324x-72x^2+4x^3. \end{aligned}$$

The tin actually allows only $$0<x<9$$ (each cut must be positive and cannot exceed half the side).

Step 1 Find the critical points

Differentiate:

$$V'(x)=\dfrac{dV}{dx}=324-144x+12x^2.$$

Set $$V'(x)=0$$:

$$\begin{aligned} 324-144x+12x^2&=0\\ 12\,(27-12x+x^2)&=0\\ x^2-12x+27&=0. \end{aligned}$$

Factorising, $$(x-3)(x-9)=0,$$ giving possible critical points $$x=3$$ cm or $$x=9$$ cm. Since $$x=9$$ would eliminate the base, only $$x=3$$ cm lies in the feasible interval.

Step 2 Second-derivative test

Compute $$V''(x)$$:

$$V''(x)=\dfrac{d^2V}{dx^2}=-144+24x.$$

At $$x=3$$, $$V''(3)=-144+72=-72<0,$$ which confirms a local maximum.

Because the only stationary point in $$0<x<9$$ is a maximum and the end-points give zero volume, this maximum is the absolute maximum.

Conclusion: the side of each small square to be cut off for maximum volume is $$3\,\text{cm}$$.

A rectangular sheet of tin measures $$45\,\mathrm{cm} \times 24\,\mathrm{cm}$$. Squares of side $$x\,\mathrm{cm}$$ are cut from each corner and the flaps folded up to form an open box. Then

• height of the box $$= x$$,
• length $$= 45 - 2x$$,
• breadth $$= 24 - 2x$$.

Volume:

$$V(x) = x(45 - 2x)(24 - 2x).$$

Expanding,

$$(45 - 2x)(24 - 2x) = 1080 - 90x - 48x + 4x^2 = 1080 - 138x + 4x^2,$$

so

$$V(x) = x(1080 - 138x + 4x^2) = 1080x - 138x^2 + 4x^3.$$

Domain. We need $$45 - 2x \gt 0$$, $$24 - 2x \gt 0$$, and $$x \gt 0$$. The binding condition is $$24 - 2x \gt 0$$, i.e. $$x \lt 12$$. Hence the admissible range is $$0 \lt x \lt 12$$.

Step 1 — Critical points.

$$V'(x) = 1080 - 276x + 12x^2 = 12(x^2 - 23x + 90).$$

Set $$V'(x) = 0$$: $$x^2 - 23x + 90 = 0$$. The discriminant is $$\Delta = 23^2 - 4(90) = 529 - 360 = 169 = 13^2$$, so

$$x = \dfrac{23 \pm 13}{2} = 18 \ \text{or}\ 5.$$

Only $$x = 5$$ lies in the admissible interval $$(0, 12)$$.

Step 2 — Second-derivative test.

$$V''(x) = -276 + 24x.$$

At $$x = 5$$, $$V''(5) = -276 + 120 = -156 \lt 0$$, so $$V$$ attains a maximum at $$x = 5$$.

Step 3 — Maximum volume.

$$V(5) = 1080(5) - 138(25) + 4(125) = 5400 - 3450 + 500 = 2450\,\mathrm{cm^3}.$$

Therefore the side of each square to be cut off is $$5\,\mathrm{cm}$$, giving a maximum volume of $$2450\,\mathrm{cm^3}$$.

Let the fixed circle have centre $$O$$ and radius $$r$$. Let $$ABCD$$ be a rectangle inscribed in this circle, with sides $$x$$ and $$y$$.

Step 1 — Relation between the sides.

Each diagonal of the rectangle is a diameter of the circle, so by Pythagoras's theorem applied to a right triangle formed by two adjacent sides and the diagonal,

$$x^2 + y^2 = (2r)^2 = 4r^2.$$

Hence $$y = \sqrt{4r^2 - x^2}$$ for $$0 < x < 2r$$.

Step 2 — Area as a function of $$x$$.

$$A(x) = xy = x\sqrt{4r^2 - x^2}, \qquad 0 < x < 2r.$$

Step 3 — Critical point.

Differentiate:

$$A'(x) = \sqrt{4r^2 - x^2} + x \cdot \dfrac{-x}{\sqrt{4r^2 - x^2}} = \dfrac{(4r^2 - x^2) - x^2}{\sqrt{4r^2 - x^2}} = \dfrac{4r^2 - 2x^2}{\sqrt{4r^2 - x^2}}.$$

Set $$A'(x) = 0$$: $$4r^2 - 2x^2 = 0 \Rightarrow x^2 = 2r^2 \Rightarrow x = \sqrt{2}\,r$$.

The corresponding $$y$$ is

$$y = \sqrt{4r^2 - 2r^2} = \sqrt{2}\,r = x,$$

so the rectangle is a square.

Step 4 — Verify it is a maximum.

• For $$0 < x < \sqrt{2}\,r$$: $$4r^2 - 2x^2 > 0$$, so $$A'(x) > 0$$.
• For $$\sqrt{2}\,r < x < 2r$$: $$4r^2 - 2x^2 < 0$$, so $$A'(x) < 0$$.

The derivative changes from positive to negative, so $$A$$ attains a maximum at $$x = \sqrt{2}\,r$$.

Step 5 — Maximum area.

$$A_{\max} = (\sqrt{2}\,r)(\sqrt{2}\,r) = 2r^2.$$

Hence among all rectangles inscribed in a given circle of radius $$r$$, the square (with side $$\sqrt{2}\,r$$) has the maximum area, namely $$2r^2$$. Proved.

Let a right circular cylinder have radius $$r \gt 0$$ and height $$h \gt 0$$. The total surface area (closed cylinder) is

$$S = 2\pi r h + 2\pi r^2 = 2\pi r(h + r).$$

Since $$S$$ is prescribed (a positive constant), this gives the constraint

$$h = \dfrac{S}{2\pi r} - r.$$

Volume to maximise.

$$V = \pi r^2 h = \pi r^2 \left(\dfrac{S}{2\pi r} - r\right) = \dfrac{Sr}{2} - \pi r^3.$$

Step 1 — Critical point.

$$V'(r) = \dfrac{S}{2} - 3\pi r^2.$$

Set $$V'(r) = 0$$:

$$\dfrac{S}{2} - 3\pi r^2 = 0 \quad\Longrightarrow\quad r^2 = \dfrac{S}{6\pi}.$$

Step 2 — Second-derivative test.

$$V''(r) = -6\pi r.$$

For every $$r \gt 0$$, we have $$V''(r) = -6\pi r \lt 0$$. By the second-derivative test, $$V$$ attains its maximum at this critical point.

Step 3 — Corresponding height.

From $$r^2 = \dfrac{S}{6\pi}$$ we obtain $$S = 6\pi r^2$$. Substituting into the constraint,

$$h = \dfrac{S}{2\pi r} - r = \dfrac{6\pi r^2}{2\pi r} - r = 3r - r = 2r.$$

Therefore $$h = 2r$$, i.e. the height equals the diameter of the base. Hence proved.

Let the required right circular cylinder have radius $$r \text{ cm}$$ and height $$h \text{ cm}$$.

Step 1 : Constraint from the volume
The can must hold $$100 \text{ cm}^3$$, so

$$V = \pi r^2 h = 100.$$

Hence \[h = \dfrac{100}{\pi r^{2}}.\tag{1}\]

Step 2 : Surface area to be minimised
For a closed cylinder,

$$S = 2\pi r h + 2\pi r^{2} = 2\pi r(h+r).$$

Substituting (1),

\[S(r) = 2\pi r\left(\dfrac{100}{\pi r^{2}} + r\right) = \dfrac{200}{r} + 2\pi r^{2}.\tag{2}\]

Step 3 : First-derivative test

$$\dfrac{dS}{dr} = -\dfrac{200}{r^{2}} + 4\pi r.$$

Set $$\dfrac{dS}{dr}=0$$ for extreme values:

$$-\dfrac{200}{r^{2}} + 4\pi r = 0 \;\Longrightarrow\; 4\pi r^{3} = 200 \;\Longrightarrow\; r^{3} = \dfrac{50}{\pi}.$$

Thus

\[r = \left(\dfrac{50}{\pi}\right)^{1/3}\;\text{cm}.\tag{3}\]

Step 4 : Second-derivative test

$$\dfrac{d^{2}S}{dr^{2}} = \dfrac{400}{r^{3}} + 4\pi.$$

Because $$r>0$$, we have $$\dfrac{d^{2}S}{dr^{2}} > 0$$, so the surface area is minimum at the critical point (3).

Step 5 : Corresponding height

From (1) and (3):

$$h = \dfrac{100}{\pi r^{2}} = \dfrac{100}{\pi\left(\dfrac{50}{\pi}\right)^{2/3}} = 2\left(\dfrac{50}{\pi}\right)^{1/3}\;\text{cm} = 2r.$$

Conclusion
For the given volume, the can with the smallest surface area has

• radius $$r = \bigl(50/\pi\bigr)^{1/3}\,\text{cm},$$
• height $$h = 2\bigl(50/\pi\bigr)^{1/3}\,\text{cm}$$  (i.e. $$h = 2r$$).

Step 1  Let the cut lengths be $$s$$ metres for the square and $$28-s$$ metres for the circle.

Conditions for each figure

• Square: perimeter $$=s \;\Rightarrow\;$$ side length $$a=\dfrac{s}{4}$$, so area $$A_\text{sq}=\left(\dfrac{s}{4}\right)^2=\dfrac{s^2}{16}$$.
• Circle: circumference $$=28-s$$, hence radius $$r=\dfrac{28-s}{2\pi}$$ and area $$A_\text{cir}=\pi r^2=\pi\left(\dfrac{28-s}{2\pi}\right)^2=\dfrac{(28-s)^2}{4\pi}$$.

Step 2  Form the objective function

Total area

$$A(s)=A_\text{sq}+A_\text{cir}=\dfrac{s^2}{16}+\dfrac{(28-s)^2}{4\pi},\qquad 0\le s\le 28.$$

Step 3  Differentiate and find critical point

$$\frac{dA}{ds}=\frac{2s}{16}+\frac{2(28-s)(-1)}{4\pi}=\frac{s}{8}-\frac{28-s}{2\pi}.$$

Set $$\dfrac{dA}{ds}=0$$:

$$\frac{s}{8}=\frac{28-s}{2\pi}\;\Rightarrow\;\pi s=4(28-s)\;\Rightarrow\;(\pi+4)s=112\;\Rightarrow\;s=\frac{112}{\pi+4}.$$

Step 4  Second-derivative test

$$\frac{d^2A}{ds^2}=\frac{1}{8}+\frac{1}{2\pi}>0,$$ so the critical point gives a minimum total area.

Step 5  Required lengths

• Length for the square: $$s=\dfrac{112}{\pi+4}\;\text{metres}\;\approx 15.68\;\text{m}.$$
• Length for the circle: $$28-s=28-\dfrac{112}{\pi+4}=\dfrac{28\pi}{\pi+4}\;\text{metres}\;\approx 12.32\;\text{m}.$$

Thus cutting the wire into pieces of the above lengths minimises the combined area of the square and the circle.

Let the sphere have centre $$O$$ and radius $$R$$. Inscribe a right circular cone with axis along a diameter of the sphere, vertex $$A$$ at the top of this diameter ($$OA = R$$), and base parallel to the equatorial plane through the point $$B$$ on the diameter. Set

• $$OB = k$$ (a signed length along the axis, with $$-R \le k \le R$$, where positive values place $$B$$ above $$O$$ — i.e. between $$O$$ and $$A$$),
• radius of the cone's base $$= r$$,
• height of the cone $$= h = AB = R - k$$ (measured from $$B$$ upward to $$A$$).

From the right triangle joining $$O$$, $$B$$ and a point on the rim of the base,

$$r^2 = R^2 - k^2.$$

Volume of the cone:

$$V(k) = \dfrac{1}{3}\pi r^2 h = \dfrac{\pi}{3}(R^2 - k^2)(R - k), \qquad -R \le k \le R.$$

Drop the positive constant factor $$\dfrac{\pi}{3}$$ and study

$$f(k) = (R^2 - k^2)(R - k) = (R + k)(R - k)^2.$$

Step 1 — First derivative.

By the product rule,

$$f'(k) = (1)(R - k)^2 + (R + k) \cdot 2(R - k)(-1) = (R - k)\bigl[(R - k) - 2(R + k)\bigr] = (R - k)(-R - 3k) = -(R - k)(R + 3k).$$

Step 2 — Critical points.

$$f'(k) = 0 \quad\Longrightarrow\quad k = R \ \text{or}\ k = -\dfrac{R}{3}.$$

At $$k = R$$ the cone degenerates ($$h = 0$$), so the only meaningful interior critical point is $$k = -\dfrac{R}{3}$$.

Step 3 — Second-derivative test.

Expand $$f'(k)$$:

$$f'(k) = -(R - k)(R + 3k) = -(R^2 + 2Rk - 3k^2).$$

Differentiate again:

$$f''(k) = -(2R - 6k) = 6k - 2R.$$

At $$k = -\dfrac{R}{3}$$,

$$f''\!\left(-\dfrac{R}{3}\right) = 6\!\left(-\dfrac{R}{3}\right) - 2R = -2R - 2R = -4R \lt 0,$$

so $$f$$ (and hence $$V$$) attains a maximum at $$k = -\dfrac{R}{3}$$.

Step 4 — Dimensions of the largest cone.

$$r^2 = R^2 - \dfrac{R^2}{9} = \dfrac{8R^2}{9}, \qquad h = R - \left(-\dfrac{R}{3}\right) = \dfrac{4R}{3}.$$

Step 5 — Maximum volume.

$$V_{\max} = \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3}\pi \cdot \dfrac{8R^2}{9} \cdot \dfrac{4R}{3} = \dfrac{32\pi R^3}{81}.$$

The volume of the sphere is $$V_{\text{sphere}} = \dfrac{4}{3}\pi R^3$$, so

$$\dfrac{V_{\max}}{V_{\text{sphere}}} = \dfrac{32\pi R^3/81}{4\pi R^3/3} = \dfrac{32}{81} \cdot \dfrac{3}{4} = \dfrac{8}{27}.$$

Hence the largest cone inscribed in a sphere of radius $$R$$ has volume $$\dfrac{8}{27}$$ that of the sphere. Proved.

Let  the radius of the base be $$r$$ and the altitude (height) be $$h$$.

Step 1. Write the formulae for volume and curved surface area.

• Volume is fixed at $$V$$: $$V = \tfrac13\,\pi r^2 h$$
• Curved surface area to be minimised: $$S = \pi r l, \;\; l = \sqrt{r^2 + h^2}$$ hence $$S = \pi r\sqrt{r^2 + h^2}$$

Step 2. Express $$h$$ in terms of $$r$$ using the volume constraint.

From $$V = \tfrac13\,\pi r^2 h$$ we get $$h = \frac{3V}{\pi r^2}$$

Step 3. Write the surface area $$S$$ as a function of $$r$$.

$$S(r) = \pi r\sqrt{\,r^2 + \Big( \tfrac{3V}{\pi r^2}\Big)^2}\;=\;\pi r\sqrt{r^2 + \frac{9V^2}{\pi^2 r^4}}$$

To simplify differentiation, pull the square-root denominator outside:

$$S(r)=\frac{\sqrt{\pi^2 r^6+9V^2}}{r}\qquad(1)$$

Step 4. Differentiate and find critical points.

Let $$N(r)=\sqrt{\pi^2 r^6+9V^2}$$. Then from (1) $$S(r)=r^{-1}N(r).$$

Differentiate w.r.t. $$r$$ using the product rule and $$N'$$:

$$N'(r)=\frac12(\pi^2 r^6+9V^2)^{-\tfrac12}\,(6\pi^2 r^5) =\frac{3\pi^2 r^5}{N(r)}.$$

$$\begin{aligned} S'(r) &= -r^{-2}N(r)+r^{-1}N'(r)\\ &= -\frac{N}{r^2}+\frac{1}{r}\cdot\frac{3\pi^2 r^5}{N} = -\frac{N}{r^2}+\frac{3\pi^2 r^4}{N}. \end{aligned}$$

Set $$S'(r)=0$$ for an extremum:

$$\frac{3\pi^2 r^4}{N}=\frac{N}{r^2}\;\;\Longrightarrow\;\;3\pi^2 r^6=N^2.$$

But $$N^2=\pi^2 r^6+9V^2$$, so

$$3\pi^2 r^6=\pi^2 r^6+9V^2\;\;\Longrightarrow\;\;2\pi^2 r^6=9V^2.$$

$$r^6=\frac{9V^2}{2\pi^2}\qquad(2)$$

Step 5. Obtain the relation between $$h$$ and $$r$$.

From the expression for $$h$$ in Step 2, compute $$\frac{h^2}{r^2}=\frac{\big(3V/\pi r^2\big)^2}{r^2}=\frac{9V^2}{\pi^2 r^6}.$$

Substitute $$r^6$$ from (2): $$\frac{h^2}{r^2}=\frac{9V^2}{\pi^2}\;\cdot\;\frac{2\pi^2}{9V^2}=2.$$

Therefore $$h^2=2r^2\;\;\Longrightarrow\;\;h=\sqrt2\,r.$$ (The positive root is chosen because $$h,r>0$$.)

Step 6. Confirm that this critical point gives a minimum.

• As $$r\to0^+$$, $$S(r)\to\infty$$ (because of the factor $$1/r$$ in (1)).
• As $$r\to\infty$$, $$S(r)\to\infty$$ (since the numerator grows like $$r^3$$).

Hence the finite stationary value found above is the least possible curved surface area.

Result: For a fixed volume, a right circular cone has the minimum curved surface area when its altitude satisfies $$h = \sqrt2\,r$$, i.e. the height equals $$\sqrt{2}$$ times the base radius.

For a right circular cone with slant height $$l$$ (given, constant), base radius $$r$$, height $$h$$, and semi-vertical angle $$\theta$$ (the angle between the axis and a generator),

$$r = l\sin\theta, \qquad h = l\cos\theta, \qquad 0 \lt \theta \lt \dfrac{\pi}{2}.$$

The volume of the cone is

$$V = \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3}\pi (l\sin\theta)^2 (l\cos\theta) = \dfrac{\pi l^3}{3}\sin^2\theta\,\cos\theta.$$

Since $$\dfrac{\pi l^3}{3}$$ is a positive constant, maximising $$V$$ is equivalent to maximising

$$f(\theta) = \sin^2\theta\,\cos\theta, \qquad 0 \lt \theta \lt \dfrac{\pi}{2}.$$

Step 1 — Critical point.

Using the product rule,

$$f'(\theta) = 2\sin\theta\cos\theta \cdot \cos\theta + \sin^2\theta \cdot (-\sin\theta) = 2\sin\theta\cos^2\theta - \sin^3\theta = \sin\theta\bigl(2\cos^2\theta - \sin^2\theta\bigr).$$

Setting $$f'(\theta) = 0$$ and noting that $$\sin\theta \ne 0$$ in $$\left(0, \dfrac{\pi}{2}\right)$$,

$$2\cos^2\theta - \sin^2\theta = 0.$$

Use $$\cos^2\theta = 1 - \sin^2\theta$$:

$$2(1 - \sin^2\theta) - \sin^2\theta = 0 \quad\Longrightarrow\quad 2 - 3\sin^2\theta = 0 \quad\Longrightarrow\quad \sin^2\theta = \dfrac{2}{3}.$$

Then $$\cos^2\theta = \dfrac{1}{3}$$ and

$$\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \sqrt{\dfrac{2/3}{1/3}} = \sqrt{2}.$$

Step 2 — Confirm it is a maximum (first-derivative sign test).

Recall $$f'(\theta) = \sin\theta\,(2\cos^2\theta - \sin^2\theta)$$. With $$\sin\theta \gt 0$$ throughout $$\left(0, \dfrac{\pi}{2}\right)$$, the sign of $$f'$$ follows that of $$g(\theta) = 2\cos^2\theta - \sin^2\theta = 2 - 3\sin^2\theta$$.

• For $$\sin^2\theta \lt \dfrac{2}{3}$$ (i.e. $$\theta \lt \tan^{-1}\sqrt{2}$$), $$g(\theta) \gt 0$$, so $$f'(\theta) \gt 0$$ — $$f$$ is increasing.
• For $$\sin^2\theta \gt \dfrac{2}{3}$$ (i.e. $$\theta \gt \tan^{-1}\sqrt{2}$$), $$g(\theta) \lt 0$$, so $$f'(\theta) \lt 0$$ — $$f$$ is decreasing.

The sign of $$f'$$ changes from positive to negative at $$\tan\theta = \sqrt{2}$$, so $$f$$ (and hence $$V$$) attains its maximum there.

Conclusion. The semi-vertical angle of the cone of maximum volume for a given slant height is $$\theta = \tan^{-1}\sqrt{2}$$. Proved.

Let a right circular cone have base radius $$r$$, slant height $$l$$, height $$h$$, and semi-vertical angle $$\theta$$, where

$$\sin\theta = \dfrac{r}{l}, \qquad \cos\theta = \dfrac{h}{l}, \qquad \tan\theta = \dfrac{r}{h}.$$

The total surface area (curved + base) is given to be constant:

$$S = \pi r l + \pi r^2.$$

The volume is

$$V = \dfrac{1}{3}\pi r^2 h.$$

Step 1 — Express the variables via $$\theta$$.

From $$\sin\theta = r/l$$, $$l = r/\sin\theta$$. Substituting into the surface-area equation,

$$S = \pi r \cdot \dfrac{r}{\sin\theta} + \pi r^2 = \pi r^2\!\left(\dfrac{1}{\sin\theta} + 1\right) = \dfrac{\pi r^2 (1 + \sin\theta)}{\sin\theta}.$$

Solving for $$r^2$$:

$$r^2 = \dfrac{S\sin\theta}{\pi (1 + \sin\theta)}. \qquad (1)$$

Also, $$h = l\cos\theta = \dfrac{r\cos\theta}{\sin\theta} = r\cot\theta$$. Hence

$$V = \dfrac{1}{3}\pi r^2 \cdot r\cot\theta = \dfrac{1}{3}\pi r^3 \cot\theta.$$

So

$$V^2 = \dfrac{\pi^2}{9}\, r^6 \cot^2\theta = \dfrac{\pi^2}{9} \cdot (r^2)^3 \cdot \dfrac{\cos^2\theta}{\sin^2\theta}.$$

Using (1),

$$V^2 = \dfrac{\pi^2}{9} \cdot \dfrac{S^3 \sin^3\theta}{\pi^3 (1 + \sin\theta)^3} \cdot \dfrac{\cos^2\theta}{\sin^2\theta} = \dfrac{S^3}{9\pi} \cdot \dfrac{\sin\theta\,\cos^2\theta}{(1 + \sin\theta)^3} = \dfrac{S^3}{9\pi} \cdot \dfrac{\sin\theta (1 - \sin^2\theta)}{(1 + \sin\theta)^3}.$$

Since $$1 - \sin^2\theta = (1 - \sin\theta)(1 + \sin\theta)$$,

$$V^2 = \dfrac{S^3}{9\pi} \cdot \dfrac{\sin\theta (1 - \sin\theta)}{(1 + \sin\theta)^2}.$$

Let $$s = \sin\theta$$, $$0 < s < 1$$. Maximising $$V$$ is equivalent to maximising

$$F(s) = \dfrac{s(1 - s)}{(1 + s)^2}.$$

Step 2 — Critical point.

$$F(s) = \dfrac{s - s^2}{(1 + s)^2}.$$

Differentiate using the quotient rule:

$$F'(s) = \dfrac{(1 - 2s)(1 + s)^2 - (s - s^2)\cdot 2(1 + s)}{(1 + s)^4} = \dfrac{(1 - 2s)(1 + s) - 2(s - s^2)}{(1 + s)^3}.$$

Expand the numerator:

$$(1 - 2s)(1 + s) = 1 - s - 2s^2, \qquad 2(s - s^2) = 2s - 2s^2.$$

So the numerator is

$$1 - s - 2s^2 - 2s + 2s^2 = 1 - 3s.$$

Hence

$$F'(s) = \dfrac{1 - 3s}{(1 + s)^3}.$$

Setting $$F'(s) = 0$$ gives $$s = \dfrac{1}{3}$$.

Step 3 — Confirm a maximum.

For $$0 < s < \dfrac{1}{3}$$, $$1 - 3s > 0$$ and $$(1 + s)^3 > 0$$, so $$F'(s) > 0$$ — $$F$$ is increasing.
For $$\dfrac{1}{3} < s < 1$$, $$1 - 3s < 0$$, so $$F'(s) < 0$$ — $$F$$ is decreasing.

The sign of $$F'$$ changes from positive to negative at $$s = \dfrac{1}{3}$$, so $$F$$ (and hence $$V$$) attains its maximum there.

Conclusion. $$\sin\theta = \dfrac{1}{3}$$, i.e. $$\theta = \sin^{-1}\!\left(\dfrac{1}{3}\right)$$. Proved.

The displacement of the car from the starting point P after $$t$$ seconds is

$$x(t)=t^{2}\Bigl(2-\frac{t}{3}\Bigr)=2t^{2}-\frac{t^{3}}{3}.$$

To find when the car stops at Q we need the instant at which its velocity becomes zero.

Step 1 · Velocity function
Velocity is the first derivative of displacement with respect to time:

$$v(t)=\frac{dx}{dt}=\frac{d}{dt}\Bigl(2t^{2}-\frac{t^{3}}{3}\Bigr)=4t-t^{2}.$$

Step 2 · Stopping time
The car stops when $$v(t)=0$$:

$$4t-t^{2}=0 \;\Longrightarrow\; t(4-t)=0.$$

This gives $$t=0$$ (the start) or $$t=4\;\text{s}$$. Hence the car reaches Q at

$$t=4\;\text{seconds}.$$

Step 3 · Distance PQ
Substitute $$t=4$$ in the displacement function:

$$x(4)=2(4)^{2}-\frac{(4)^{3}}{3}=32-\frac{64}{3}=\frac{96-64}{3}=\frac{32}{3}\;\text{m}.$$

Therefore the distance between P and Q is

$$\displaystyle \frac{32}{3}\;\text{metres}\;\bigl(10\dfrac{2}{3}\,\text{m}\bigr).$$

Let the depth (height) of water at time $$t$$ be $$h\text{ m}$$ and the corresponding radius of the water surface be $$r\text{ m}$$.

The semi-vertical angle of the cone is $$\alpha$$ with $$\tan\alpha = 0.5$$, so

$$\tan\alpha = \dfrac{r}{h} \;\Longrightarrow\; r = 0.5h$$.

The volume $$V$$ of water in the cone when the depth is $$h$$ is

$$V = \dfrac{1}{3}\pi r^{2}h = \dfrac{1}{3}\pi (0.5h)^{2}h = \dfrac{\pi}{12}h^{3}.$$

Water is poured in at the constant rate

$$\frac{dV}{dt}=5\;\text{m}^{3}\text{h}^{-1}.$$

Differentiating the volume relation with respect to time $$t$$,

$$\frac{dV}{dt}=\frac{\pi}{12}\cdot 3h^{2}\frac{dh}{dt}=\frac{\pi}{4}h^{2}\frac{dh}{dt}.$$

Hence

$$\frac{dh}{dt}=\frac{4}{\pi h^{2}}\,\frac{dV}{dt}.$$

At the instant when $$h = 4\text{ m}$$,

$$\frac{dh}{dt}=\frac{4}{\pi(4)^{2}}\times 5 = \frac{5}{4\pi}\;\text{m h}^{-1}\;\approx 0.398\;\text{m h}^{-1}.$$

Therefore, the level of the water is rising at $$\dfrac{5}{4\pi}\,\text{m h}^{-1}$$ when the depth of the water is 4 m.

Step 1 : Notation

Let

• the foot of the lamp-post be the origin O.
• the man’s position at time $$t$$ be the point M so that $$OM = x\text{ m}$$.
• the tip of the shadow be the point S so that $$MS = y\text{ m}$$.

All distances are measured along the ground and are taken positive away from the lamp-post. Thus the total distance from O to S is $$OS = x + y$$.

Step 2 : A geometric relation between $$x$$ and $$y$$

The lamp-post OR and its shadow OS form a right-triangle similar to the right-triangle formed by the man’s height MP and his shadow MS:

$$\dfrac{\text{height of lamp}}{\text{length of its shadow}} = \dfrac{\text{height of man}}{\text{length of his shadow}} \;\;\Longrightarrow\;\; \dfrac{6}{x+y} = \dfrac{2}{y}.$$

Cross-multiplying and simplifying,

$$6y = 2(x + y) \;\Longrightarrow\; 6y = 2x + 2y \;\Longrightarrow\; 4y = 2x \;\Longrightarrow\; y = \dfrac{x}{2}. $$

Step 3 : Differentiate with respect to time

$$y = \dfrac{x}{2} \;\Longrightarrow\; \dfrac{dy}{dt} = \dfrac{1}{2}\,\dfrac{dx}{dt}.$$

Step 4 : Insert the given walking speed

The man walks away from the lamp at a uniform speed of $$5\,\text{km h}^{-1}$$.

Since we are measuring $$x$$ and $$y$$ in metres, convert the speed to metres per hour:

$$\dfrac{dx}{dt} = 5\,\text{km h}^{-1} = 5\times1000 = 5000\,\text{m h}^{-1}.$$

Therefore

$$\dfrac{dy}{dt} = \dfrac{1}{2}\times5000 = 2500\,\text{m h}^{-1}.$$

In kilometres per hour,

$$\dfrac{dy}{dt} = \dfrac{2500}{1000} = 2.5\,\text{km h}^{-1}.$$

For completeness, in metres per second,

$$\dfrac{dy}{dt} = 2.5\times\dfrac{1000}{3600} = \dfrac{25}{36}\,\text{m s}^{-1}\;\,(\approx0.694\,\text{m s}^{-1}).$$

Conclusion

The length of the man’s shadow is increasing at $$2.5\,\text{km h}^{-1}$$ (about $$0.694\,\text{m s}^{-1}$$).

Let $$f(x) = \dfrac{3}{10}x^4 - \dfrac{4}{5}x^3 - 3x^2 + \dfrac{36}{5}x + 11$$.

Step 1 — First derivative.

$$f'(x) = \dfrac{3}{10}(4x^3) - \dfrac{4}{5}(3x^2) - 6x + \dfrac{36}{5} = \dfrac{6}{5}x^3 - \dfrac{12}{5}x^2 - 6x + \dfrac{36}{5}.$$

Factor out $$\dfrac{6}{5}$$:

$$f'(x) = \dfrac{6}{5}\bigl(x^3 - 2x^2 - 5x + 6\bigr).$$

Step 2 — Critical points.

Solve $$x^3 - 2x^2 - 5x + 6 = 0$$. Try $$x = 1$$: $$1 - 2 - 5 + 6 = 0$$, so $$x = 1$$ is a root. Dividing,

$$x^3 - 2x^2 - 5x + 6 = (x - 1)(x^2 - x - 6) = (x - 1)(x - 3)(x + 2).$$

Hence

$$f'(x) = \dfrac{6}{5}(x + 2)(x - 1)(x - 3),$$

and $$f'(x) = 0$$ at $$x = -2,\ 1,\ 3$$. These split the real line into four sub-intervals: $$(-\infty, -2)$$, $$(-2, 1)$$, $$(1, 3)$$, and $$(3, \infty)$$.

Step 3 — Sign of $$f'(x)$$ on each sub-interval.

The factor $$\dfrac{6}{5}$$ is positive, so the sign of $$f'(x)$$ equals the sign of $$(x + 2)(x - 1)(x - 3)$$.

• On $$(-\infty, -2)$$, test $$x = -3$$: $$(x + 2)(x - 1)(x - 3) = (-1)(-4)(-6) = -24 \lt 0$$, so $$f'(x) \lt 0$$. Hence $$f$$ is decreasing on $$(-\infty, -2)$$.
• On $$(-2, 1)$$, test $$x = 0$$: $$(0 + 2)(0 - 1)(0 - 3) = (2)(-1)(-3) = 6 \gt 0$$, so $$f'(x) \gt 0$$. Hence $$f$$ is increasing on $$(-2, 1)$$.
• On $$(1, 3)$$, test $$x = 2$$: $$(2 + 2)(2 - 1)(2 - 3) = (4)(1)(-1) = -4 \lt 0$$, so $$f'(x) \lt 0$$. Hence $$f$$ is decreasing on $$(1, 3)$$.
• On $$(3, \infty)$$, test $$x = 4$$: $$(4 + 2)(4 - 1)(4 - 3) = (6)(3)(1) = 18 \gt 0$$, so $$f'(x) \gt 0$$. Hence $$f$$ is increasing on $$(3, \infty)$$.

Step 4 — Conclusion.

• (a) $$f$$ is increasing on $$(-2, 1) \cup (3, \infty)$$.
• (b) $$f$$ is decreasing on $$(-\infty, -2) \cup (1, 3)$$.

We show that $$f(x) = \tan^{-1}(\sin x + \cos x)$$ is strictly increasing on $$\left(0, \dfrac{\pi}{4}\right)$$.

Step 1 — Differentiate.

Let $$u(x) = \sin x + \cos x$$, so $$f(x) = \tan^{-1} u(x)$$. By the chain rule,

$$f'(x) = \dfrac{u'(x)}{1 + u(x)^2}, \qquad u'(x) = \cos x - \sin x,$$

hence

$$f'(x) = \dfrac{\cos x - \sin x}{1 + (\sin x + \cos x)^2}.$$

Step 2 — Sign of the denominator.

$$1 + (\sin x + \cos x)^2 \gt 0 \quad \text{for every real } x,$$

so the sign of $$f'(x)$$ is determined entirely by the numerator $$\cos x - \sin x$$.

Step 3 — Sign of the numerator on $$\left(0, \dfrac{\pi}{4}\right)$$.

For $$0 \lt x \lt \dfrac{\pi}{4}$$:

• $$\sin x$$ is strictly increasing on $$\left[0, \dfrac{\pi}{4}\right]$$, so $$\sin x \lt \sin\dfrac{\pi}{4} = \dfrac{\sqrt{2}}{2}$$.
• $$\cos x$$ is strictly decreasing on $$\left[0, \dfrac{\pi}{4}\right]$$, so $$\cos x \gt \cos\dfrac{\pi}{4} = \dfrac{\sqrt{2}}{2}$$.

Chaining these inequalities gives

$$\sin x \lt \dfrac{\sqrt{2}}{2} \lt \cos x,$$

so

$$\cos x - \sin x \gt 0 \quad \text{for every } x \in \left(0, \dfrac{\pi}{4}\right).$$

Step 4 — Conclusion.

Since both the numerator and the denominator of $$f'(x)$$ are positive on $$\left(0, \dfrac{\pi}{4}\right)$$,

$$f'(x) \gt 0 \quad \text{for every } x \in \left(0, \dfrac{\pi}{4}\right),$$

so $$f$$ is strictly increasing on this interval. Hence proved.

Given data

• Let the radius at time t seconds be $$r\,(\text{cm}).$$
• The rate of increase of the radius is $$\dfrac{dr}{dt}=0.05\,\text{cm s}^{-1}.$$
• We have to find $$\dfrac{dA}{dt}$$ when $$r=3.2\,\text{cm}.$$

Relation between area and radius

The area $$A$$ of a circle of radius $$r$$ is given by $$A=\pi r^{2}.$$

Differentiating with respect to time

Using the chain rule,

$$\dfrac{dA}{dt}=\dfrac{d}{dt}(\pi r^{2})=\pi\cdot2r\cdot\dfrac{dr}{dt}=2\pi r\dfrac{dr}{dt}.$$

Substituting the known values

When $$r=3.2\,\text{cm}$$ and $$\dfrac{dr}{dt}=0.05\,\text{cm s}^{-1},$$

$$\dfrac{dA}{dt}=2\pi(3.2)(0.05)=0.32\pi\,\text{cm}^{2}\text{s}^{-1}.$$

Numerical value (optional)

Taking $$\pi\approx3.14,$$ we get

$$\dfrac{dA}{dt}\approx0.32\times3.14\approx1.005\,\text{cm}^{2}\text{s}^{-1}.$$

Hence, the area of the disc is increasing at the rate of about $$1.0\,\text{cm}^{2}\text{s}^{-1}$$ when its radius is $$3.2\,\text{cm}.$$

Step 1 : Model the problem

Let the side length of the square cut from each corner be $$x\text{ m}$$ ( $$x>0$$ ).
After removing the four squares and folding the flaps upward:

• Height of the box = $$x\text{ m}$$.
• New length = $$8-2x\text{ m}$$ (because a strip of $$x$$ metres is removed from both long edges).
• New width = $$3-2x\text{ m}$$ (because a strip of $$x$$ metres is removed from both short edges).

Thus the volume (in m³) is

$$V(x)=x\,(8-2x)\,(3-2x).$$

Step 2 : Write $$V(x)$$ in polynomial form

First multiply the last two factors:

$$(8-2x)(3-2x)=24-22x+4x^{2}.$$

Hence

$$V(x)=x\,(24-22x+4x^{2})=24x-22x^{2}+4x^{3}.$$

Step 3 : Determine the domain of $$x$$

Both $$8-2x$$ and $$3-2x$$ must be positive:

$$8-2x>0\;\Longrightarrow\;x<4,$$
$$3-2x>0\;\Longrightarrow\;x<1.5.$$

Therefore the physically meaningful interval is $$0<x<1.5.$$ (We also exclude $$x=0$$ because the box would be flat.)

Step 4 : Locate the critical points

Differentiate $$V(x)$$:

$$V'(x)=\frac{d}{dx}(24x-22x^{2}+4x^{3})=24-44x+12x^{2}.$$

Set $$V'(x)=0$$ to find stationary points:

$$12x^{2}-44x+24=0\;\Longrightarrow\;3x^{2}-11x+6=0.$$

Using the quadratic formula,

$$x=\frac{11\pm\sqrt{11^{2}-4\cdot3\cdot6}}{2\cdot3}=\frac{11\pm\sqrt{49}}{6}=\frac{11\pm7}{6}.$$

This gives two roots:

• $$x_{1}=\dfrac{2}{3}\text{ m}=0.666\dots\text{ m}$$,
• $$x_{2}=3\text{ m}$$ (outside the domain $$0<x<1.5$$, so it is discarded).

Thus the only critical point in the domain is $$x=\dfrac{2}{3}\text{ m}$$.

Step 5 : Confirm that it yields a maximum

Compute the second derivative:

$$V''(x)=\frac{d}{dx}(12x^{2}-44x+24)=24x-44.$$

Evaluate at $$x=\dfrac{2}{3}:$$

$$V''\!\left(\tfrac{2}{3}\right)=24\left(\tfrac{2}{3}\right)-44=16-44=-28<0.$$

Because $$V''(x)<0$$ at this point, the volume is maximum when $$x=\dfrac{2}{3}\text{ m}$$.

Step 6 : Compute the maximum volume

Substitute $$x=\dfrac{2}{3}$$ into $$V(x)=24x-22x^{2}+4x^{3}$$:

\[\begin{aligned} V_{\max}&=24\left(\tfrac{2}{3}\right)-22\left(\tfrac{2}{3}\right)^{2}+4\left(\tfrac{2}{3}\right)^{3}\\[2pt] &=16-\frac{88}{9}+\frac{32}{27}\\[2pt] &=\frac{432}{27}-\frac{264}{27}+\frac{32}{27}\\[2pt] &=\frac{200}{27}.\end{aligned}\]

Hence

$$V_{\max}=\dfrac{200}{27}\;\text{m}^{3}\;(\approx7.41\;\text{m}^{3}).$$

Conclusion

The largest possible open-topped box obtained from the given sheet has:

• Squares cut of side $$\dfrac{2}{3}\text{ m}$$ from each corner,
• Maximum volume $$\dfrac{200}{27}\text{ m}^{3}\;\bigl(\approx7.41\text{ m}^{3}\bigr).$$

Let $$x$$ be the number of items produced and sold.

1. Revenue (selling price × quantity)

Selling price per item = $$5-\dfrac{x}{100}$$ rupees.

Hence revenue $$R(x)=x\left(5-\frac{x}{100}\right)=5x-\frac{x^{2}}{100}$$ rupees.

2. Total cost

Given cost $$C(x)=\frac{x}{5}+500$$ rupees.

3. Profit function

Profit $$P(x)=R(x)-C(x)$$.

Therefore
$$P(x)=\Bigl(5x-\frac{x^{2}}{100}\Bigr)-\Bigl(\frac{x}{5}+500\Bigr)=5x-\frac{x^{2}}{100}-\frac{x}{5}-500.$$

Simplifying the linear terms:

$$5x-\frac{x}{5}=\frac{25x}{5}-\frac{x}{5}=\frac{24x}{5}.$$

Thus
$$P(x)=-\frac{x^{2}}{100}+\frac{24x}{5}-500.$$

4. Maximising the profit

Differentiate:

$$P'(x)=-\frac{2x}{100}+\frac{24}{5}=-\frac{x}{50}+\frac{24}{5}.$$

Put $$P'(x)=0$$ for a stationary point:

$$-\frac{x}{50}+\frac{24}{5}=0\;\;\Longrightarrow\;\;\frac{x}{50}=\frac{24}{5}\;\;\Longrightarrow\;\;x=50\times\frac{24}{5}=240.$$

Second derivative:
$$P''(x)=-\frac{1}{50}<0$$ for all $$x$$, so the stationary point gives a maximum.

5. Verification of feasibility

At $$x=240$$, the selling price is $$5-\frac{240}{100}=2.6$$ rupees > 0, hence the value is admissible.

Conclusion: The manufacturer should sell 240 items to earn the maximum possible profit.

Given function: $$f(x)=\frac{\log x}{x}$$, where the logarithm is natural.
Domain: $$x>0$$ (because $$\log x$$ is defined only for positive real numbers).

1. First derivative

Write $$f(x)=\log x\,\cdot x^{-1}$$ and apply the product rule (or the quotient rule):

$$f'(x)=\bigl(\,\dfrac{1}{x}\bigr)\,x^{-1}+\log x\,(-x^{-2}) = \dfrac{1-\log x}{x^{2}}.$$

2. Critical points

Set $$f'(x)=0$$ (numerator must vanish because the denominator $$x^{2}>0$$):

$$1-\log x = 0 \;\Longrightarrow\; \log x = 1 \;\Longrightarrow\; x = e.$$

Thus $$x=e$$ is the only critical point in the domain $$x>0$$.

3. Second derivative test

Differentiating $$f'(x)=\bigl(1-\log x\bigr)x^{-2}$$ once more:

$$\begin{aligned} f''(x) & = \left(-\dfrac{1}{x}\right)x^{-2} + \bigl(1-\log x\bigr)(-2)x^{-3}\\[2pt] & = -x^{-3} -2\bigl(1-\log x\bigr)x^{-3}\\[2pt] & = \dfrac{-1-2\bigl(1-\log x\bigr)}{x^{3}}\\[2pt] & = \dfrac{-1-2+2\log x}{x^{3}} = \dfrac{2\log x-3}{x^{3}}. \end{aligned}$$

Evaluate at the critical point $$x=e$$:

$$f''(e)=\dfrac{2\log e-3}{e^{3}}=\dfrac{2(1)-3}{e^{3}}=\dfrac{-1}{e^{3}} < 0.$$

Since the second derivative is negative, $$f$$ is concave downward at $$x=e$$. Hence $$x=e$$ gives a local maxima.

4. Nature of the maximum

Because $$f'(x)>0$$ for $$0<x<e$$ (numerator $$1-\log x>0$$) and $$f'(x)<0$$ for $$x>e$$ (numerator $$1-\log x<0$$), the function rises up to $$x=e$$ and falls thereafter; therefore this local maximum is indeed the absolute (global) maximum of $$f(x)$$ on its domain.

5. Maximum value

Though not asked, one may note $$f(e)=\dfrac{\log e}{e}=\dfrac{1}{e}.$$

Thus, the function $$f(x)=\dfrac{\log x}{x}$$ attains its maximum at $$x=e$$.

Let the equal sides of the isosceles triangle have length $$x\text{ cm}$$ at time $$t\text{ s}$$. The base is fixed at $$b\text{ cm}$$. We are given

• $$\dfrac{dx}{dt} = -3\;\text{cm s}^{-1}$$ (the equal sides are shrinking).
• The required quantity is $$\dfrac{dA}{dt}$$ when $$x = b$$.

Describe the triangle: Draw an isosceles triangle with base $$b$$ on the bottom. Drop a perpendicular from the vertex opposite the base to the midpoint of the base; this is the height $$h$$. Each half–base is $$b/2$$, forming two congruent right-angled triangles with hypotenuse $$x$$.

1. Express the height in terms of $$x$$.

The Pythagoras theorem on one of the right triangles gives

$$h^2 = x^2 - \left(\dfrac{b}{2}\right)^2 = x^2 - \dfrac{b^2}{4}$$

so that

$$h = \sqrt{x^2 - \dfrac{b^2}{4}}.$$

2. Write the area $$A$$ as a function of $$x$$.

The area of a triangle is one-half the product of its base and height, therefore

$$A = \dfrac12\,b\,h = \dfrac12\,b\,\sqrt{x^2 - \dfrac{b^2}{4}}.$$

3. Differentiate both sides with respect to time $$t$$ (remember, $$b$$ is constant):

$$\begin{aligned} \dfrac{dA}{dt} &= \dfrac12\,b\; \dfrac{d}{dt}\Bigl(\sqrt{x^2 - \dfrac{b^2}{4}}\Bigr)\\[4pt] &= \dfrac12\,b\; \dfrac{1}{2\sqrt{x^2 - \dfrac{b^2}{4}}}\;(2x)\;\dfrac{dx}{dt}\\[4pt] &= \dfrac{b\,x}{2\,\sqrt{x^2 - \dfrac{b^2}{4}}}\;\dfrac{dx}{dt}. \end{aligned}$$

4. Evaluate at the required instant $$x = b$$.

First compute the denominator:

$$\sqrt{x^2 - \dfrac{b^2}{4}}\Bigm|_{x=b} = \sqrt{b^2 - \dfrac{b^2}{4}} = \sqrt{\dfrac{3b^2}{4}} = \dfrac{b\sqrt3}{2}.$$

Now substitute $$x=b$$, $$\dfrac{dx}{dt} = -3\text{ cm s}^{-1}$$:

$$\begin{aligned} \dfrac{dA}{dt}\Bigm|_{x=b} &= \dfrac{b\,(b)}{2\,\dfrac{b\sqrt3}{2}}\;(-3)\\[4pt] &= \dfrac{b^2}{\dfrac{b\sqrt3}{1}}\;(-3)\\[4pt] &= -\dfrac{3b^2}{b\sqrt3}\\[4pt] &= -\dfrac{3b}{\sqrt3} = -b\sqrt3\;\text{cm}^2\text{ s}^{-1}. \end{aligned}$$

5. Interpretation: The negative sign shows the area is decreasing. Hence, when the two equal sides equal the base, the area is decreasing at the rate

$$b\sqrt3\;\text{cm}^2\text{ per second}.$$

Let $$f(x) = \dfrac{4\sin x - 2x - x\cos x}{2 + \cos x}$$.

Step 1 — Domain.

Since $$-1 \le \cos x \le 1$$, $$2 + \cos x \ge 1 > 0$$, so the denominator never vanishes; the domain of $$f$$ is all of $$\mathbf{R}$$.

Step 2 — Differentiate.

Let $$N(x) = 4\sin x - 2x - x\cos x$$ and $$D(x) = 2 + \cos x$$. Then

$$N'(x) = 4\cos x - 2 - (\cos x - x\sin x) = 3\cos x - 2 + x\sin x,$$

$$D'(x) = -\sin x.$$

By the quotient rule,

$$f'(x) = \dfrac{N'(x)D(x) - N(x)D'(x)}{(2 + \cos x)^2}.$$

Step 3 — Simplify the numerator.

First term:

$$N'(x)D(x) = (3\cos x - 2 + x\sin x)(2 + \cos x).$$

Expand:

$$= 6\cos x + 3\cos^2 x - 4 - 2\cos x + 2x\sin x + x\sin x\cos x$$

$$= 3\cos^2 x + 4\cos x - 4 + 2x\sin x + x\sin x\cos x.$$

Second term:

$$-N(x)D'(x) = -(4\sin x - 2x - x\cos x)(-\sin x) = (4\sin x - 2x - x\cos x)\sin x$$

$$= 4\sin^2 x - 2x\sin x - x\sin x\cos x.$$

Adding the two:

$$3\cos^2 x + 4\cos x - 4 + 2x\sin x + x\sin x\cos x + 4\sin^2 x - 2x\sin x - x\sin x\cos x$$

$$= 3\cos^2 x + 4\sin^2 x + 4\cos x - 4.$$

Using $$\sin^2 x = 1 - \cos^2 x$$,

$$3\cos^2 x + 4(1 - \cos^2 x) + 4\cos x - 4 = -\cos^2 x + 4\cos x = \cos x\,(4 - \cos x).$$

Hence

$$f'(x) = \dfrac{\cos x\,(4 - \cos x)}{(2 + \cos x)^2}.$$

Step 4 — Sign of $$f'(x)$$.

• $$(2 + \cos x)^2 > 0$$ for every $$x$$.
• $$4 - \cos x \ge 4 - 1 = 3 > 0$$ for every $$x$$.

Therefore the sign of $$f'(x)$$ equals the sign of $$\cos x$$. That is,

• $$f'(x) > 0$$ whenever $$\cos x > 0$$, i.e. on $$\left(2k\pi - \dfrac{\pi}{2},\ 2k\pi + \dfrac{\pi}{2}\right)$$ for integer $$k$$;
• $$f'(x) < 0$$ whenever $$\cos x < 0$$, i.e. on $$\left(2k\pi + \dfrac{\pi}{2},\ 2k\pi + \dfrac{3\pi}{2}\right)$$ for integer $$k$$.

Conclusion.

• (i) $$f$$ is increasing on $$\left(2k\pi - \dfrac{\pi}{2},\ 2k\pi + \dfrac{\pi}{2}\right)$$ for every $$k \in \mathbf{Z}$$.
• (ii) $$f$$ is decreasing on $$\left(2k\pi + \dfrac{\pi}{2},\ 2k\pi + \dfrac{3\pi}{2}\right)$$ for every $$k \in \mathbf{Z}$$.

Setup. Take the ellipse $$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$$ with $$a \gt b \gt 0$$. Place one vertex of the triangle at $$A(a, 0)$$, the right end of the major axis. By symmetry, the other two vertices may be taken as $$P(x, y)$$ and $$Q(x, -y)$$ on the ellipse, with $$y \gt 0$$ and $$-a \le x \lt a$$.

Step 1 — Area of $$\triangle APQ$$.

The base $$PQ$$ is vertical with length $$2y$$, and the perpendicular distance from $$A(a, 0)$$ to the line $$PQ$$ (the vertical line $$X = x$$) is $$a - x$$. Hence

$$\Delta = \dfrac{1}{2}\,(2y)\,(a - x) = y(a - x).$$

Since $$P$$ lies on the ellipse, $$y = b\sqrt{1 - \dfrac{x^2}{a^2}}$$. Therefore

$$\Delta(x) = b\sqrt{1 - \dfrac{x^2}{a^2}}\,(a - x), \qquad -a \lt x \lt a.$$

Step 2 — Substitution.

Let $$x = at$$ with $$-1 \lt t \lt 1$$. Then

$$\Delta(t) = ab\,(1 - t)\sqrt{1 - t^2}.$$

Since $$\Delta \gt 0$$, maximising $$\Delta$$ is equivalent to maximising the function

$$g(t) = (1 - t)^2(1 - t^2),$$

because $$g(t) = \dfrac{\Delta(t)^2}{a^2 b^2}$$, and the square is order-preserving on non-negative quantities. Factor $$1 - t^2 = (1 - t)(1 + t)$$ to get

$$g(t) = (1 - t)^3 (1 + t).$$

Step 3 — Critical points of $$g(t)$$.

By the product rule applied to $$g(t) = (1 - t)^3 (1 + t)$$,

$$g'(t) = -3(1 - t)^2(1 + t) + (1 - t)^3 = (1 - t)^2\bigl[-3(1 + t) + (1 - t)\bigr] = (1 - t)^2(-2 - 4t) = -2(1 - t)^2(1 + 2t).$$

Set $$g'(t) = 0$$: either $$(1 - t)^2 = 0$$ (which gives $$t = 1$$, corresponding to a degenerate triangle with $$\Delta = 0$$) or $$1 + 2t = 0$$, giving $$t = -\dfrac{1}{2}$$.

Step 4 — Confirm a maximum.

The factor $$(1 - t)^2 \ge 0$$ always, so the sign of $$g'(t) = -2(1 - t)^2(1 + 2t)$$ is the sign of $$-(1 + 2t)$$.

• For $$-1 \lt t \lt -\dfrac{1}{2}$$: $$1 + 2t \lt 0$$, so $$g'(t) \gt 0$$ — $$g$$ is increasing.
• For $$-\dfrac{1}{2} \lt t \lt 1$$: $$1 + 2t \gt 0$$, so $$g'(t) \lt 0$$ — $$g$$ is decreasing.

Hence $$g$$ (and therefore $$\Delta$$) attains its maximum at $$t = -\dfrac{1}{2}$$.

Step 5 — Maximum area.

At $$t = -\dfrac{1}{2}$$, $$x = -\dfrac{a}{2}$$ and

$$y = b\sqrt{1 - \dfrac{1}{4}} = \dfrac{b\sqrt{3}}{2}.$$

Therefore

$$\Delta_{\max} = y(a - x) = \dfrac{b\sqrt{3}}{2} \cdot \dfrac{3a}{2} = \dfrac{3\sqrt{3}}{4}\,ab.$$

Let the rectangular base of the tank have length $$x\,\mathrm{m}$$ and breadth $$y\,\mathrm{m}$$, with fixed depth $$h = 2\,\mathrm{m}$$.

Step 1 — Volume constraint.

$$V = xyh = 8 \quad\Rightarrow\quad xy = \dfrac{8}{2} = 4.$$

Step 2 — Cost of the base.

Area of base $$= xy = 4\,\mathrm{m^2}$$. At Rs $$70$$/m²,

$$C_{\text{base}} = 70 \times 4 = \text{Rs } 280.$$

Step 3 — Cost of the four sides.

Two sides have area $$xh = 2x$$ each, and the other two have area $$yh = 2y$$ each. So the total side area is

$$2(xh) + 2(yh) = 2x \cdot 2 + 2y \cdot 2 = 4x + 4y \quad \mathrm{m^2}.$$

At Rs $$45$$/m²,

$$C_{\text{sides}} = 45(4x + 4y) = 180(x + y).$$

Step 4 — Total cost as a function of $$x$$.

$$C(x, y) = 280 + 180(x + y).$$

Using $$y = \dfrac{4}{x}$$ ($$x > 0$$),

$$C(x) = 280 + 180\!\left(x + \dfrac{4}{x}\right).$$

Step 5 — Minimise $$C(x)$$.

$$C'(x) = 180\!\left(1 - \dfrac{4}{x^2}\right).$$

Set $$C'(x) = 0$$: $$1 - \dfrac{4}{x^2} = 0 \Rightarrow x^2 = 4 \Rightarrow x = 2$$ (taking $$x > 0$$). Then $$y = \dfrac{4}{2} = 2$$.

Step 6 — Second-derivative test.

$$C''(x) = 180 \cdot \dfrac{8}{x^3} = \dfrac{1440}{x^3} > 0 \ \text{for}\ x > 0,$$

so $$x = 2$$ gives the minimum cost.

Step 7 — Minimum cost.

$$C_{\min} = 280 + 180(2 + 2) = 280 + 720 = \text{Rs } 1000.$$

The least-cost tank has base $$2\,\mathrm{m} \times 2\,\mathrm{m}$$ and depth $$2\,\mathrm{m}$$, costing Rs $$1000$$.

Let the radius of the circle be $$r$$ and the side of the square be $$s$$.

Given condition (perimeter)

$$2\pi r + 4s = k \;\;\;\; (1)$$

From (1) we express $$s$$ in terms of $$r$$:

$$s = \dfrac{k - 2\pi r}{4} \;\;\;\; (2)$$

Total area

$$A(r) = \pi r^2 + s^2$$

Substitute (2):

$$A(r) = \pi r^2 + \left(\dfrac{k - 2\pi r}{4}\right)^2$$

Simplify the second term:

$$A(r) = \pi r^2 + \dfrac{(k - 2\pi r)^2}{16} \;\;\;\; (3)$$

Step 1: Differentiate

$$\dfrac{\mathrm dA}{\mathrm dr} = 2\pi r + \dfrac{1}{16}\,2(k-2\pi r)(-2\pi)$$

$$\dfrac{\mathrm dA}{\mathrm dr} = 2\pi r - \dfrac{\pi}{4}(k - 2\pi r)$$

Step 2: Stationary point

Set $$\dfrac{\mathrm dA}{\mathrm dr}=0$$:

$$2\pi r = \dfrac{\pi}{4}(k - 2\pi r)$$

Multiply by $$\dfrac{4}{\pi}$$:

$$8r = k - 2\pi r$$

$$8r + 2\pi r = k$$

$$r = \dfrac{k}{8 + 2\pi} \;\;\;\; (4)$$

Step 3: Corresponding value of s

Using (2) and (4):

$$s = \dfrac{k - 2\pi\dfrac{k}{8 + 2\pi}}{4} = \dfrac{k}{4}\left(1 - \dfrac{2\pi}{8 + 2\pi}\right)$$

$$s = \dfrac{k}{4}\left(\dfrac{8}{8 + 2\pi}\right) = \dfrac{2k}{8 + 2\pi} \;\;\;\; (5)$$

Comparing (4) and (5):

$$s = 2r$$

Step 4: Second derivative test

Differentiate $$\dfrac{\mathrm dA}{\mathrm dr}$$ once more:

$$\dfrac{\mathrm d^2A}{\mathrm dr^2} = 2\pi - \dfrac{\pi}{4}(-2\pi) = 2\pi + \dfrac{\pi^2}{2} > 0$$

Since the second derivative is positive, $$A(r)$$ attains its minimum at the stationary point.

Conclusion

The sum of the areas of the circle and the square is least when the side of the square equals twice the radius of the circle, i.e.

$$s = 2r.$$

Let the rectangular part of the window have width $$2r\,\mathrm{m}$$ (the diameter of the semicircle on top) and height $$h\,\mathrm{m}$$, where $$r > 0$$ is the radius of the semicircle.

Step 1 — Perimeter constraint.

The total perimeter consists of the two vertical sides of the rectangle ($$2h$$), the base of the rectangle ($$2r$$), and the curved edge of the semicircle ($$\pi r$$). So

$$2h + 2r + \pi r = 10 \quad\Rightarrow\quad h = \dfrac{10 - r(2 + \pi)}{2}.$$

For $$h \ge 0$$, $$r \le \dfrac{10}{2 + \pi}$$.

Step 2 — Area to maximise.

The light-admitting area is

$$A = (2r)(h) + \dfrac{1}{2}\pi r^2 = 2rh + \dfrac{1}{2}\pi r^2.$$

Substitute the expression for $$h$$:

$$A(r) = 2r \cdot \dfrac{10 - r(2 + \pi)}{2} + \dfrac{1}{2}\pi r^2 = r\bigl[10 - r(2 + \pi)\bigr] + \dfrac{1}{2}\pi r^2.$$

Simplify:

$$A(r) = 10r - (2 + \pi)r^2 + \dfrac{\pi}{2}r^2 = 10r - \!\left(2 + \dfrac{\pi}{2}\right)r^2.$$

Step 3 — First derivative.

$$A'(r) = 10 - 2\!\left(2 + \dfrac{\pi}{2}\right)r = 10 - (4 + \pi)r.$$

Set $$A'(r) = 0$$: $$(4 + \pi)r = 10 \Rightarrow r = \dfrac{10}{4 + \pi}$$.

Step 4 — Second-derivative test.

$$A''(r) = -(4 + \pi) < 0,$$

so $$A$$ attains a maximum at $$r = \dfrac{10}{4 + \pi}$$.

Step 5 — Dimensions at maximum area.

Radius of semicircle: $$r = \dfrac{10}{4 + \pi}\,\mathrm{m}$$.

Height of rectangle:

$$h = \dfrac{10 - r(2 + \pi)}{2} = \dfrac{10}{2} - \dfrac{r(2 + \pi)}{2}.$$

Compute $$r(2 + \pi) = \dfrac{10(2 + \pi)}{4 + \pi}$$, so

$$h = 5 - \dfrac{5(2 + \pi)}{4 + \pi} = \dfrac{5(4 + \pi) - 5(2 + \pi)}{4 + \pi} = \dfrac{5 \cdot 2}{4 + \pi} = \dfrac{10}{4 + \pi}\,\mathrm{m} = r.$$

Width of rectangle (and diameter of semicircle): $$2r = \dfrac{20}{4 + \pi}\,\mathrm{m}$$.

Let the right–angled triangle be placed in the coordinate plane with the right angle at the origin O, one leg on the positive x-axis and the other on the positive y-axis.

• Let $$A(c,0)$$ be the point on the x-axis.
• Let $$B(0,d)$$ be the point on the y-axis.
• The hypotenuse is the line segment $$AB$$.
• A point $$P(b,a)$$ is taken on the hypotenuse so that its perpendicular distances from the two axes (i.e. the legs) are $$b$$ and $$a$$ respectively.

The equation of the hypotenuse through $$A(c,0)$$ and $$B(0,d)$$ is

$$y=-\dfrac{d}{c}x+d$$.

Since $$P(b,a)$$ lies on this line, we have

$$a=-\dfrac{d}{c}b+d$$  ⇔  $$d=a+\dfrac{d}{c}b$$.

Put $$k=\dfrac{d}{c}>0$$ (slope magnitude). Then $$d=kc$$ and the above relation becomes

$$a=k(c-b).$$

Thus

$$c=\dfrac{a}{k}+b \qquad (1)$$

and $$d=kc.$$

The length of the hypotenuse is

$$H=\sqrt{c^{2}+d^{2}}=c\sqrt{1+k^{2}}.$$

Substituting (1), we obtain the length as a function of the single positive variable $$k$$:

$$H(k)=\left(\dfrac{a}{k}+b\right)\sqrt{1+k^{2}}.$$

Minimisation

It is more convenient to minimise $$H^{2}$$:

$$f(k)=H^{2}(k)=\left(\dfrac{a}{k}+b\right)^{2}(1+k^{2}),\qquad k>0.$$

Differentiate:

Let $$u=\dfrac{a}{k}+b\;\;(\Rightarrow\;\dfrac{du}{dk}=-\dfrac{a}{k^{2}}).$$

Then

$$\dfrac{df}{dk}=2u\dfrac{du}{dk}(1+k^{2})+u^{2}(2k)=0.$$

Dividing by $$2u\;(u\neq0)$$ gives

$$-\dfrac{a}{k^{2}}(1+k^{2})+uk=0\quad\Longrightarrow\quad u k^{3}=a(1+k^{2}).$$

Substituting $$u=\dfrac{a}{k}+b$$:

$$\left(\dfrac{a}{k}+b\right)k^{3}=a(1+k^{2})$$

$$\Rightarrow\;ak^{2}+bk^{3}=a+ak^{2}$$

$$\Rightarrow\;bk^{3}=a.$$

Hence the critical value is

$$k=\left(\dfrac{a}{b}\right)^{1/3}.\qquad (2)$$

As $$f(k)\to\infty$$ when $$k\to0^{+}$$ or $$k\to\infty$$ and there is a unique critical point, (2) gives the point of minimum hypotenuse.

Value of the minimum hypotenuse

Write $$a=x^{3},\;b=y^{3}$$ so that $$x=a^{1/3},\;y=b^{1/3}$$.

From (1) and (2):

$$c=\dfrac{a}{k}+b=a^{2/3}b^{1/3}+b=y(x^{2}+y^{2}).$$

Also

$$1+k^{2}=1+\left(\dfrac{x}{y}\right)^{2}=\dfrac{x^{2}+y^{2}}{y^{2}}$$

and hence $$\sqrt{1+k^{2}}=\dfrac{\sqrt{x^{2}+y^{2}}}{y}.$$

Therefore

$$H_{\min}=c\sqrt{1+k^{2}}=y(x^{2}+y^{2})\cdot\dfrac{\sqrt{x^{2}+y^{2}}}{y}=(x^{2}+y^{2})^{3/2}.$$

Returning to $$a,b$$:

$$H_{\min}=\left(a^{2/3}+b^{2/3}\right)^{3/2}.$$

Thus the minimum possible length of the hypotenuse is exactly

$$\boxed{\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}}.\quad\blacksquare$$

Let $$f(x) = \cos^2 x + \sin x$$ on the closed interval $$[0, \pi]$$.

Step 1 — Critical points.

$$f'(x) = 2\cos x \cdot (-\sin x) + \cos x = -2\sin x\cos x + \cos x = \cos x\,(1 - 2\sin x).$$

Set $$f'(x) = 0$$:

• $$\cos x = 0 \Rightarrow x = \dfrac{\pi}{2}$$ (in $$[0, \pi]$$);
• $$1 - 2\sin x = 0 \Rightarrow \sin x = \dfrac{1}{2} \Rightarrow x = \dfrac{\pi}{6}\ \text{or}\ \dfrac{5\pi}{6}$$ (both in $$[0, \pi]$$).

Critical points in $$[0, \pi]$$: $$x = \dfrac{\pi}{6},\ \dfrac{\pi}{2},\ \dfrac{5\pi}{6}$$.

Step 2 — Function values at the critical points and the endpoints.

$$f(0) = \cos^2 0 + \sin 0 = 1 + 0 = 1.$$

$$f\!\left(\dfrac{\pi}{6}\right) = \left(\dfrac{\sqrt{3}}{2}\right)^2 + \dfrac{1}{2} = \dfrac{3}{4} + \dfrac{1}{2} = \dfrac{5}{4}.$$

$$f\!\left(\dfrac{\pi}{2}\right) = 0 + 1 = 1.$$

$$f\!\left(\dfrac{5\pi}{6}\right) = \left(-\dfrac{\sqrt{3}}{2}\right)^2 + \dfrac{1}{2} = \dfrac{3}{4} + \dfrac{1}{2} = \dfrac{5}{4}.$$

$$f(\pi) = (-1)^2 + 0 = 1.$$

Step 3 — Absolute extrema.

• Absolute maximum $$= \dfrac{5}{4}$$, attained at $$x = \dfrac{\pi}{6}$$ and $$x = \dfrac{5\pi}{6}$$.
• Absolute minimum $$= 1$$, attained at $$x = 0,\ \dfrac{\pi}{2},\ \pi$$.

Let a sphere have centre $$O$$ and radius $$r$$. Inscribe a right circular cone with its axis along a diameter, vertex $$A$$ on the sphere (the upper end of the diameter), and base centre $$B$$ on the same diameter, with the base circle in the plane perpendicular to the diameter through $$B$$.

Let

• $$OB = y$$ be the distance from the sphere's centre to the base plane (measured along the axis in the direction opposite to $$A$$), with $$0 \le y \lt r$$;
• $$x$$ be the radius of the base of the cone;
• $$h = AB = AO + OB = r + y$$ be the altitude of the cone.

Because the rim of the base lies on the sphere, the right triangle with hypotenuse $$r$$ (the line from $$O$$ to a rim point) and legs $$x$$ and $$y$$ gives

$$x^2 + y^2 = r^2 \quad\Longrightarrow\quad x^2 = r^2 - y^2.$$

Step 1 — Volume as a function of $$y$$.

$$V(y) = \dfrac{1}{3}\pi x^2 h = \dfrac{\pi}{3}(r^2 - y^2)(r + y), \qquad 0 \le y \lt r.$$

Step 2 — Find the critical point.

Differentiate:

$$V'(y) = \dfrac{\pi}{3}\bigl[(r^2 - y^2)(1) + (r + y)(-2y)\bigr] = \dfrac{\pi}{3}\bigl[r^2 - y^2 - 2ry - 2y^2\bigr] = \dfrac{\pi}{3}\bigl[r^2 - 2ry - 3y^2\bigr].$$

Set $$V'(y) = 0$$: $$3y^2 + 2ry - r^2 = 0$$. By the quadratic formula,

$$y = \dfrac{-2r \pm \sqrt{4r^2 + 12r^2}}{6} = \dfrac{-2r \pm 4r}{6}.$$

Taking the positive root, $$y = \dfrac{2r}{6} = \dfrac{r}{3}$$ (the other root $$y = -r$$ is inadmissible).

Step 3 — Second-derivative test.

$$V''(y) = \dfrac{\pi}{3}(-2r - 6y).$$

At $$y = \dfrac{r}{3}$$,

$$V''\!\left(\dfrac{r}{3}\right) = \dfrac{\pi}{3}\!\left(-2r - 6 \cdot \dfrac{r}{3}\right) = \dfrac{\pi}{3}(-2r - 2r) = -\dfrac{4\pi r}{3} \lt 0,$$

so $$V$$ attains a maximum at $$y = \dfrac{r}{3}$$.

Step 4 — Required altitude.

$$h = r + y = r + \dfrac{r}{3} = \dfrac{4r}{3}.$$

Hence the altitude of the right circular cone of maximum volume inscribed in a sphere of radius $$r$$ is $$\dfrac{4r}{3}$$. Proved.

Given. A function $$f$$ defined on $$[a, b]$$ such that $$f'(x) \gt 0$$ for every $$x \in (a, b)$$. In particular, $$f$$ is differentiable on $$(a, b)$$, and (since differentiability implies continuity) it is also continuous on $$(a, b)$$.

To prove. $$f$$ is strictly increasing on $$(a, b)$$, i.e. for any $$x_1, x_2 \in (a, b)$$ with $$x_1 \lt x_2$$, we have $$f(x_1) \lt f(x_2)$$.

Step 1 — Apply the Mean Value Theorem.

Choose any $$x_1, x_2 \in (a, b)$$ with $$x_1 \lt x_2$$. Then $$[x_1, x_2] \subset (a, b)$$. By hypothesis $$f$$ is differentiable on $$(a, b)$$, hence:

• $$f$$ is continuous on the closed interval $$[x_1, x_2]$$,
• $$f$$ is differentiable on the open interval $$(x_1, x_2)$$.

So the Mean Value Theorem applies: there exists $$c \in (x_1, x_2)$$ such that

$$\dfrac{f(x_2) - f(x_1)}{x_2 - x_1} = f'(c).$$

Step 2 — Use the sign of $$f'(c)$$.

Since $$c \in (x_1, x_2) \subset (a, b)$$, by hypothesis $$f'(c) \gt 0$$; and $$x_2 - x_1 \gt 0$$. Therefore

$$f(x_2) - f(x_1) = f'(c)\,(x_2 - x_1) \gt 0,$$

i.e. $$f(x_1) \lt f(x_2)$$.

Step 3 — Conclusion.

The pair $$x_1 \lt x_2$$ in $$(a, b)$$ was arbitrary, so $$f(x_1) \lt f(x_2)$$ for every such pair. Hence $$f$$ is strictly increasing on $$(a, b)$$. Proved.

Inscribe a right circular cylinder in a sphere of radius $$R$$ with axis along a diameter of the sphere. Let the radius of the cylinder be $$r$$ and its height be $$h$$, with $$0 < h < 2R$$. By symmetry, each circular end of the cylinder is at distance $$\dfrac{h}{2}$$ from the centre of the sphere. The Pythagorean theorem applied to the right triangle joining the sphere's centre, the centre of an end face, and a point on the rim gives

$$r^2 + \left(\dfrac{h}{2}\right)^2 = R^2 \quad\Rightarrow\quad r^2 = R^2 - \dfrac{h^2}{4}.$$

The volume of the cylinder is

$$V = \pi r^2 h = \pi h\left(R^2 - \dfrac{h^2}{4}\right) = \pi\left(R^2 h - \dfrac{h^3}{4}\right), \qquad 0 < h < 2R.$$

Step 1 — First derivative.

$$\dfrac{dV}{dh} = \pi\left(R^2 - \dfrac{3h^2}{4}\right).$$

Set $$\dfrac{dV}{dh} = 0$$:

$$R^2 - \dfrac{3h^2}{4} = 0 \quad\Rightarrow\quad h^2 = \dfrac{4R^2}{3} \quad\Rightarrow\quad h = \dfrac{2R}{\sqrt{3}}\ (\text{taking}\ h > 0).$$

Step 2 — Second-derivative test.

$$\dfrac{d^2 V}{dh^2} = \pi\left(-\dfrac{3h}{2}\right) = -\dfrac{3\pi h}{2}.$$

At $$h = \dfrac{2R}{\sqrt{3}}$$,

$$\dfrac{d^2 V}{dh^2} = -\dfrac{3\pi}{2} \cdot \dfrac{2R}{\sqrt{3}} = -\dfrac{3\pi R}{\sqrt{3}} = -\pi R\sqrt{3} < 0,$$

so $$V$$ attains a maximum at $$h = \dfrac{2R}{\sqrt{3}}$$.

Step 3 — Maximum volume.

$$r^2 = R^2 - \dfrac{1}{4}\left(\dfrac{2R}{\sqrt{3}}\right)^2 = R^2 - \dfrac{R^2}{3} = \dfrac{2R^2}{3}.$$

Therefore

$$V_{\max} = \pi r^2 h = \pi \cdot \dfrac{2R^2}{3} \cdot \dfrac{2R}{\sqrt{3}} = \dfrac{4\pi R^3}{3\sqrt{3}}.$$

Hence the height of the cylinder of maximum volume is $$\dfrac{2R}{\sqrt{3}}$$ and its maximum volume is $$\dfrac{4\pi R^3}{3\sqrt{3}}$$.

Consider a right circular cone of height $$h$$ and semi-vertical angle $$\alpha$$. Inscribe a cylinder coaxial with the cone, with its lower base on the base of the cone and its upper face inside the cone. Let the cylinder's height be $$x$$ (measured upward from the base), so its top face lies at distance $$h - x$$ from the apex.

At that height, the radius of the cone's cross-section equals the radius $$r$$ of the cylinder. From similar right triangles (axis, generator, horizontal radius at the apex-distance $$h - x$$),

$$\dfrac{r}{h - x} = \tan\alpha \quad\Longrightarrow\quad r = (h - x)\tan\alpha.$$

Hence the volume of the cylinder is

$$V(x) = \pi r^2 x = \pi \tan^2\alpha\,(h - x)^2 x, \qquad 0 \le x \le h.$$

Let $$k = \pi \tan^2\alpha \gt 0$$, so that $$V(x) = k\,x(h - x)^2$$.

Step 1 — First derivative.

$$V'(x) = k\bigl[(h - x)^2 + x \cdot 2(h - x)(-1)\bigr] = k(h - x)\bigl[(h - x) - 2x\bigr] = k(h - x)(h - 3x).$$

Set $$V'(x) = 0$$: either $$h - x = 0$$ (giving $$x = h$$, which makes the cylinder degenerate with $$r = 0$$) or $$h - 3x = 0$$, giving

$$x = \dfrac{h}{3}.$$

Step 2 — Second-derivative test.

$$V''(x) = k\bigl[-(h - 3x) - 3(h - x)\bigr] = k(-h + 3x - 3h + 3x) = k(6x - 4h).$$

At $$x = \dfrac{h}{3}$$,

$$V''\!\left(\dfrac{h}{3}\right) = k\!\left(6 \cdot \dfrac{h}{3} - 4h\right) = k(2h - 4h) = -2kh \lt 0,$$

so $$V$$ attains its maximum at $$x = \dfrac{h}{3}$$.

Step 3 — Maximum volume.

At $$x = \dfrac{h}{3}$$,

$$r = (h - x)\tan\alpha = \left(h - \dfrac{h}{3}\right)\tan\alpha = \dfrac{2h}{3}\tan\alpha,$$

so

$$V_{\max} = \pi r^2 x = \pi\!\left(\dfrac{2h}{3}\tan\alpha\right)^{\!2}\!\left(\dfrac{h}{3}\right) = \pi \cdot \dfrac{4h^2}{9}\tan^2\alpha \cdot \dfrac{h}{3} = \dfrac{4\pi h^3 \tan^2\alpha}{27}.$$

Hence the height of the cylinder of greatest volume is $$\dfrac{h}{3}$$ (one-third of the cone's height), and the greatest volume is $$\dfrac{4}{27}\pi h^3 \tan^2\alpha$$. Proved.