KVPY SB SX Exam 2011 Question Paper

Question : 89 of 120

Marks: +1, -0

The maximum possible value of

x^2+y^2-4x-6y

, x, y real, subject to the condition

|x+y|+|x-y|=4

Solution:

| x + y | + | x – y | = 4 represent a square

x^2 + y^2 - 4x - 6y

= (x - 2)^2 + (y - 3)^2 -13

= (distance point on square from

(2, 3))^2 -13

Maximum

= (- 2 - 2)^2 + (- 2 - 3)^2 - 13 = 28

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