The maximum possible value of x^2+y^2-4x-6y,w | vedclass

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(B) The given condition $|x+y|+|x-y|=4$ represents a square in the $xy$-plane with vertices at $(2, 2), (-2, 2), (-2, -2), (2, -2)$ is incorrect; let

Vedclass · Accepted Answer

$28$

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(B) The given condition $|x+y|+|x-y|=4$ represents a square in the $xy$-plane with vertices at $(2, 2), (-2, 2), (-2, -2), (2, -2)$ is incorrect; let us re-evaluate.The condition $|x+y|+|x-y|=4$ is equivalent to $\max(|x|, |y|) = 2$,which represents a square with vertices at $(2, 2), (-2, 2), (-2, -2), (2, -2)$.We want to maximize $f(x, y) = x^2+y^2-4x-6y = (x-2)^2 + (y-3)^2 - 13$.This expression represents the square of the distance from the point $(2, 3)$ to any point $(x, y)$ on the square,minus $13$.To maximize this,we look for the point on the square boundary furthest from $(2, 3)$.The vertices of the square are $V_1(2, 2), V_2(-2, 2), V_3(-2, -2), V_4(2, -2)$.Calculating the squared distances $d^2$ from $(2, 3)$ to each vertex:$d_1^2 = (2-2)^2 + (2-3)^2 = 1$$d_2^2 = (-2-2)^2 + (2-3)^2 = 16 + 1 = 17$$d_3^2 = (-2-2)^2 + (-2-3)^2 = 16 + 25 = 41$$d_4^2 = (2-2)^2 + (-2-3)^2 = 25$The maximum squared distance is $41$ at vertex $(-2, -2)$.Thus,the maximum value of $f(x, y) = 41 - 13 = 28$.

The maximum possible value of $x^2+y^2-4x-6y$,where $x, y$ are real,subject to the condition $|x+y|+|x-y|=4$ is

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