Hi Kumar,

Remember that a cubic flattens at two points, $$x=\frac{5}{3}$$ is equally viable, just that it is not an integer value, $$\frac{5}{3}≈ 1.\overline{6}$$, so you can try $$x=2$$ which  is the closest integer to $$1.\overline{6}$$. 

Remember that, trying $$4$$, as it is the closest integer to $$5$$ makes sense as well, but it is further from $$5$$ than $$2$$ was from $$1.\overline{6}$$. Not only that, but even putting $$x=4$$ gives us the volume as $$16$$, so we can reject it that way also.

Hope this helps, feel free to ask if you have any doubts

Hi Kumar Simha,
You can get the answer even through maxima and minima. You obtained the value of x, but we are expected to calculate the volume. So, substitute the values obtained in the expression; you will get the answer for one of the values.
I hope this helps!!

We get x=5.
Volume you get is zero
Pls see the pic i have sent.

Hi Kumar Simha,
If you substitute x = 5/3, you get the answer. You can't put x = 5. Why? The side length of the initial cardboard is 10; if you cut 2 *5 = 10, there will be nothing to create a box with. So, you should only consider 5/3.
I hope this helps!!

Got it. If we take the double differentiation of the equation <0, to ensure its a maxima, we get 6x-20=0.... I.e., x<3.333
That way we could conclude x close to 2 as you said👍..n not,4

Hi Kumar, yes you can do that as well, feel free to ask if you have any doubts!

Hi, Thank you for reporting. We have updated the solution

Option C is obtained as answer when differentiation gives x = (5/3).
However, this violates the condition in question that x is also an integer.
Hence, differentiation cannot give us the correct answer in this case.

Hi, you can get benefit from AM>=GM only when sum or product is constant. here 10-2x+x is not constant. So, you can't apply here directly so, use the following approach

Let A=10-2x

B=2x

$$\frac{\left(\left(\frac{A}{2}+\frac{A}{2}\right)+B\right)}{3}\ge\sqrt[\ 3]{\frac{A}{2}\cdot\frac{A}{2}\cdot B}$$

=> $$\frac{\left(\left(\frac{10-2x}{2}+\frac{10-2x}{2}\right)+2x\right)}{3}\ge\sqrt[\ 3]{\frac{10-2x}{2}\cdot\frac{10-2x}{2}\cdot2x}$$

$$\frac{\left(10\right)}{3}\ge\sqrt[\ 3]{\frac{\left(10-2x\right)^2}{4}\cdot2x}$$

=> $$\left(\frac{\left(10\right)}{3}\right)^3.4\ge\ \left(10-2x\right)^2\cdot2x$$

=> $$\left(\frac{\left(10\right)}{3}\right)^3.2\ge\ \left(10-2x\right)^2\cdot x$$

So, max value is $$\left(\frac{\left(10\right)}{3}\right)^3.2\ =74.07$$

Hope this helps.