CAT 2001QA Question 36

Inequality Maximization / MinimizationEasy

Let x, y be two positive numbers such that x + y = 1. Then, the minimum value of (x+1x)2+(y+1y)2 is ______.

Answer & solution

Correct answer: 12.5

  • A

    12

  • B

    20

  • 12.5

  • D

    13.3

Solution

The sum of two numbers, when their product is known, is minimum when they are equal.

The product of two numbers, when their sum is known, is maximum when they are equal.

(x+1x)2+(y+1y)2

= (x2 + y2) + x2+y2x2y2+4

This expression is minimum when (x2 + y2) is minimum and x2y2 is maximum.

We have x + y = 1

Squaring,

x2 + 2xy + y2 = 1

∴ x2 + y2 = 1 – 2xy

x2 + y2 is minimum when xy is maximum.

xy is maximum when x = y

∴ For minimum value, both x and y have to be equal.

∴ x = y = 12

(x+1x)2+(y+1y)2=(2+12)2+(2+12)2

= (2.5)2 + (2.5)2

= 12.5

Hence, option (c).

Related Inequality Maximization / Minimization questions

See all Inequalities & Modulus questions →
CAT 2001 QA Q36: Let x, y be two positive numbers such that x + y = 1. Then, the minimum value of x + 1 x 2 + y + 1 y 2 is ____ — Solution | TheCATExam