# Maxima and Minima Questions for CAT

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## Maxima and Minima Questions for CAT:

Download important CAT Maxima and Minima Questions PDF based on previous asked questions in CAT and other MBA exams. Top 25 Maxima and Minima functions questions for CAT quantitative aptitude.

Question 1: If y is a real number, what is the difference in the maximum and minimum values obtained by $\frac{y+5}{y^2+5y+25}$ ?

a) 2/15

b) 4/15

c) 1/5

d) 1/15

Question 2: What is the maximum value of g(x) = 22 – |x+8| and for what value of x is this reached?

a) 22, 8

b) 14, 0

c) 14, -8

d) None of these

Question 3: If 2 $\leq$ x $\leq$ 3 and 4 $\leq$ y $\leq$ 5, what are the minimum and maximum values of $\frac{y+x}{y+2x}$

a) 9/10, 7/8

b) 10/11, 8/9

c) 7/10, 7/9

d) None of these

Question 4: If $x^{2} – 5x + 4 <= 0$ and $y^{2} -6y+5 <= 0$, what are the maximum and minimum values of $\frac{y+7x}{y+4x}$

a) 29/17 , 12/9

b) 23/13, 11/8

c) 25/11, 14/11

d) None of these

Question 5: If X and Y are positive real numbers and 3X+4Y = 14, what is the maximum value of $X^{3}*Y^4$ ?

a) 64

b) 128

c) 2187

d) None of these

Question 6: What is the maximum value of g(x) = |34-x| – |x + 11|?

a) 34

b) 45

c) 23

d) None of these

Question 7: If X and Y are positive and X+Y = 10, what is the maximum value of $X^3*{Y^2}$ ?

a) 3125

b) 3456

c) 3600

d) None of these

Question 8: If f(x) = min(3x-4, 11-7x). What is the maximum value of f(x)?

a) 3/2

b) -3/2

c) 1/2

d) Infinite

Question 9: For how many real values does the function $F(X) = 28 – |X^2+6|$ attain its maximum value

a) 1

b) 2

c) 3

d) None of these

Question 10: If ‘a’ and ‘b’ are two whole numbers such that |a-2|+|b-3| = 4, what is the maximum value of a*b ?

a) 24

b) 18

c) 14

d) 20

Question 11: If X,Y and Z are real numbers such that $-6 \leq X \leq -2$ and $-4 \leq Y \leq 4$ and $3 \leq Z \leq 7$. If $W=\frac{Y}{XZ}$, which of the following is necessarily true?

a) $0 \leq W \leq 2$

b) $-1/2 \leq W \leq 1/2$

c) $-2/3 \leq W \leq 1/3$

d) $-2/3 \leq W \leq 2/3$

Question 12: Two natural numbers, X and Y, satisfy the equation 2X+Y = 33. Find the maximum value of XY?

a) 135

b) 140

c) 136

d) 144

Question 13: Two natural numbers, A and B, satisfy the equation 3A+2B = 64. Find the maximum value of ${A*B}$?

a) 160

b) 162

c) 170

d) 180

Question 14: If a,b and c are three whole numbers such that |a-1|+|b-2|+|c-3|=3, what is the maximum value of (a+1)(b+2)(c+3)?

a) 120

b) 144

c) 112

d) 105

Question 15: If a,b and c are positive real numbers, what is the minimum value of a*(1/b+1/c)+b*(1/c+1/a)+c*(1/a+1/b)

a) 9

b) 6

c) 3

d) 27

Question 16: If $x^2 – 5x + 4 \leq 0$ and $y^2 -6y +5 \leq 0$, then what are the maximum and minimum values of $(y+7x)/(y+4x)$.

a) 29/17, 12/9

b) 23/13, 11/8

c) 25/11, 14/11

d) 29/11, 11/9

e) None of these

Question 17: Let a, b, c, d be four integers such that a+b+c+d = 4m+1 where m is a positive integer. Given m, which one of the following is necessarily true?

a) The minimum possible value of $a^2 + b^2 + c^2 + d^2$ is $4m^2-2m+1$

b) The minimum possible value of $a^2 + b^2 + c^2 + d^2$ is $4m^2+2m+1$

c) The maximum possible value of $a^2 + b^2 + c^2 + d^2$ is $4m^2-2m+1$

d) The maximum possible value of $a^2 + b^2 + c^2 + d^2$ is $4m^2+2m+1$

Question 18: If x, y, z are distinct positive real numbers the $(x^2(y+z) + y^2(x+z) + z^2(x+y))/xyz$ would be

a) greater than 4

b) greater than 5

c) greater than 6

d) None of the above

Question 19: If a, b, c and d are four positive real numbers such that abcd = 1, what is the minimum value of (1 + a)(1+b)(1+c)(1+d)?

a) 4

b) 1

c) 16

d) 18

Question 20: Let x and y be two positive numbers such that $x + y = 1.$ Then the minimum value of $(x+\frac{1}{x})^2+(y+\frac{1}{y})^2$ is

a) 12

b) 20

c) 12.5

d) 13.3

Question 21: If a, b, c and d are four positive real numbers such that abcd = 1, what is the minimum value of (1 + a)(1+b)(1+c)(1+d)?[CAT 2001]

a) 4

b) 1

c) 16

d) 18

Question 22: Let x and y be two positive numbers such that $x + y = 1.$Then the minimum value of $(x+\frac{1}{x})^2+(y+\frac{1}{y})^2$ is[CAT 2001]

a) 12

b) 20

c) 12.5

d) 13.3

Question 23: If a, b, c, d are positive numbers, what is the minimum value of (a+b+c+d)*(1/a+1/b+1/c+1/d)?

a) 9

b) 16

c) 25

d) 1

Question 24: If | r – 6 | = 11 and | 2q – 12 | = 8, what is the minimum possible value of q / r?

a) -2/5

b) 2/17

c) 10/17

d) None of these

Question 25: If ‘$p$’ and ‘$q$’ are two positive rational numbers such that $p+q=1$, which of the following statements is true?

a) The minimum value of $p^{\frac{1}{4}}+q^{\frac{1}{4}}$ is $2^\frac{1}{2}$

b) The maximum value of $p^{\frac{1}{4}}+q^{\frac{1}{4}}$ is $2^\frac{1}{2}$

c) The maximum value of $p^{\frac{1}{3}}+q^{\frac{1}{3}}$ is $2^\frac{2}{3}$

d) The minimum value of $p^{\frac{1}{3}}+q^{\frac{1}{3}}$ is $2^\frac{2}{3}$

Let $\frac{y+5}{y^2+5y+25} = k$

So, $ky^2+5ky+25k-y-5=0$

For the above quadriatic equation to have real roots, its discriminant should be greater than or equal to zero.

$(5k-1)^2 \geq 4*k*5*(5k-1)$

$(15k+1)(5k-1)\leq 0$

So $-1/15 \leq k \leq 1/5$

So, the difference in the maximum and minimum value is 4/15

g(x) reaches its maximum when |x+8| =0 or x = -8. So, maximum value of g(x) is 22

$\frac{y+x}{y+2x}$ = 1- $\frac{x}{y+2x}$, which equals 1 – $\frac{1}{\frac{y}{x}+2}$. So, the function is at its maximum when y/x is at its highest (x=2, y= 5) and is at its minimum when y/x is lowest (x=3, y= 4).

$\frac{y+7x}{y+4x}$ = 1 + $\frac{3x}{y+4x}$, which equals 1 – $\frac{3}{y/x+4}$. This reaches its maximum when $\frac{y}{x}$ is at its minimum and vice versa. So, the minimum value of $\frac{y+7x}{y+4x}$ is $\frac{5+7}{5+4}$ = 4/3 and the maximum is $\frac{1+28}{1+16}$ = 29/17

$X+X+X+Y+Y+Y+Y = 14.$ So,$(X+X+X+Y+Y+Y+Y)/7 \geq (X*X*X*Y*Y*Y*Y)^{1/7}$ So, $(X^{3}*Y^{4})^{1/7} \leq 2$. So, the maximum value of $X^{3}*Y^4$ is 128

If $x \leq -11$, g(x) = 45.
If $-11 < x \leq 34$, g(x) = 23-2x which peaks when x= -11.
If $x> 34$, g(x) = -45.
So, the maximum value of g(x) is 45

$X/3 + X/3 +X/3 + Y/2 + Y/2 = 10.$ So, $X^{3}*Y^{2} \leq 2^{5} * 3^{3}*2^{2} = 3456.$ The equality happens when X=6 and Y=4

f(x) reaches its maximum value when 3x-4 = 11-7x or if x = 3/2 So, the maximum value of f(x) is 1/2

F(x) attains its maximum value when X = 0. So, the number of real values for which F(X) attains its minimum value is 1

There are twelve integral solutions to the equation |a-2|+|b-3| = 4 The product a*b is maximized when a=4 and b=5 and equals 20

The minimum value of W is when Y=4, X = -2 and Z=3 and equals -2/3 The maximum value of W is when Y=-4, X=-2 and Z=3 and equals 2/3

The maximum value occurs when 2X and Y are closest to each other. So, when X=8 and Y=17, XY reaches its peak of 136.

The maximum value occurs when 3A and 2B are closest to each other. So, when A=10 and B=17, AB reaches its peak of 170.

There are fifteen solutions to the equation |a-1|+|b-2|+|c-3|=3.
Let |a-1| be x, |b-2| be y and |c-3| be z.
x, y and z are whole numbers.
x + y + z = 3.
Therefore, the number of solutions to this equation will be n+r-1 C r-1 = 3+3-1C2 = 5C2 = 10.
There are 10 solutions to the equation x+y+z = 3.
After enumerating them, we find that (a+1)(b+2)(c+3) is maximum when a=b=c=3 and its maximum value is 120.

a*(1/b+1/c)+b*(1/c+1/a)+c*(1/a+1/b) equals (a/b+b/a)+(b/c+c/b)+(a/c+c/a).
All these terms are at their minimum when a = b = c and the minimum value of the sum is 6

Option b) is the correct answer.

$\frac{y+7x}{y+4x}$ = 1 + $\frac{3x}{y+4x}$, which equals 1 + $\frac{3}{y/x+4}$.

This reaches its maximum when $\frac{y}{x}$ is at its minimum and vice versa.

So, the minimum value of $\frac{y+7x}{y+4x}$ is $\frac{5+7}{5+4}$ = 4/3 and the maximum is $\frac{1+28}{1+16}$ = 29/17

Taking lowest possible positive value of m i.e. 1 . Such that a+b+c+d=5 , so atleast one of them must be grater than 1 ,

take a=b=c=1 and d=2

we get $a^2 + b^2 + c^2 + d^2 = 7$ which is equal to $4m^2+2m+1$ for other values it is greater than $4m^2+2m+1$ . so option B

For the given expression value of x,y,z are distinct positive integers . So the value of expression will always be greater than value when all the 3 variables are equal . substitute x=y=z we get minimum value of 6 .

$(x^2(y+z) + y^2(x+z) + z^2(x+y))/xyz$ = x/z + x/y + y/z + y/x + z/y + z/x

Applying AM greater than or equal to GM, we get minimum sum = 6

Since the product is constant,

(a+b+c+d)/4 >= $(abcd)^{1/4}$

We know that abcd = 1.

Therefore, a+b+c+d >= 4

$(a+1)(b+1)(c+1)(d+1)$
= $1+a+b+c+d+ ab + ac + ad + bc + bd + cd+ abc+ bcd+ cda+ dab+abcd$
We know that $abcd = 1$
Therefore, $a = 1/bcd, b = 1/acd, c = 1/bda$ and $d = 1/abc$
Also, $cd = 1/ab, bd = 1/ac, bc = 1/ad$
The expression can be clubbed together as $1 + abcd + (a + 1/a) + (b+1/b) + (c+1/c) + (d+1/d) + (ab + 1/ab) + (ac+ 1/ac) + (ad + 1/ad)$
For any positive real number $x$, $x + 1/x \geq 2$
Therefore, the least value that $(a+1/a), (b+1/b)….(ad+1/ad)$ can take is 2.
$(a+1)(b+1)(c+1)(d+1) \geq 1 + 1 + 2 + 2 + 2+ 2 + 2 + 2 + 2$
=> $(a+1)(b+1)(c+1)(d+1) \geq 16$
The least value that the given expression can take is 16. Therefore, option C is the right answer.

Approach 1:

The given expression is symmetric in x and y and the limiting condition (x+y=1) is also symmetric in x and y.

=>This means that the expression attains the minimum value when x = y
x=y=1/2
So, the value = $(x+\frac{1}{x})^2+(y+\frac{1}{y})^2$ = $(2+\frac{1}{2})^2+(2+\frac{1}{2})^2$ =12.5

Approach 2:

$(x+1/x)^2$ + $(y+1/y)^2$ = $(x+1/x+y+1/y)^2$ – $2*(x+1/x)(y+1/y)$

Let x+1/x and y+1/y be two terms. Thus (x+1/x+y+1/y)/2 would be their Arithmetic Mean(AM) and $\sqrt{(x+1/x)(y+1/y)}$ would be their Geometric Mean (GM).

Therefore, we can express the above equation as $(x+1/x)^2$ + $(y+1/y)^2$ = $4AM^2$ – $GM^2$. As AM >= GM, the minimum value of expression would be attained when AM = GM.

When AM = GM, both terms are equal. That is x+1/x = y +1/y.

Substituting y=1-x we get

x+1/x = (1-x) + 1/(1-x)

On solving we get 2x-1 = (2x-1)/ x(1-x)

So either 2x-1 = 0 or x(1-x) = 1

x(1-x) = x * y

As x and y are positive numbers whose sum = 1, 0<= x, y <=1. Hence, their product cannot be 1.

Thus, 2x-1 = 0 or x=1/2

=> y = 1/2

So, the value = $(x+\frac{1}{x})^2+(y+\frac{1}{y})^2$ = $(2+\frac{1}{2})^2+(2+\frac{1}{2})^2$ = 12.5

Since the product is constant, the expression (1+a)(1+b)(1+c)(1+d) is least when a = b = c = d.
So, a = b = c = d = 1 and the minimum value of the expression is 2*2*2*2 = 16.

The expression attains the minimum value when x = y
x=y=1/2
So, the value = $2.5^2 * 2 = 12.5$

AM >= HM

So, (a+b+c+d)/4 >= 4/(1/a+1/b+1/c+1/d) => (a+b+c+d)*(1/a+1/b+1/c+1/d) >= 4*4 = 16

The mean of $\frac{1}{3}$ power of two numbers is less than $\frac{1}{3}$ power of the mean of the numbers.
Hence, $\frac{p^{\frac{1}{3}}+q^{\frac{1}{3}}}{2} \leq {(\frac{p+q}{2})}^{\frac{1}{3}}$
So, $p^\frac{1}{3}+q^\frac{1}{3} \leq 2^\frac{2}{3}$