Question

Find the extreme value of function x³+y³-63(x+y)+12xy

Answer 1

one doubt in the question that is x^3 is +ve only or it is -ve

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Answer 2

Answer:

the extreme value of the function is a local maximum, and it occurs at the points (-4, 3) and (3, -4).

Step-by-step explanation:

To find the extreme value of the function f(x,y) = x³+y³-63(x+y)+12xy, we need to find its critical points and then determine whether they correspond to a maximum or minimum.

To find the critical points, we need to find the partial derivatives of the function with respect to x and y and set them equal to zero.

∂f/∂x = 3x² + 12y - 63= 0

∂f/∂y = 3y² + 12x - 63 = 0

Solving these equations simultaneously gives us the critical points (x, y) = (-4, 3) and (3, -4)

To determine the nature of the extreme value at these points, we can evaluate the second partial derivatives of the function at these points. If the second partial derivative is positive at the point, it is a local minimum. If it is negative, it is a local maximum. If it is zero, it is a saddle point.

fxx(x,y) = 6x

fyy(x,y) = 6y

fxy(x,y) = 12

fxx(-4,3) = -72, fyy(-4,3) = -72, so the critical point (-4,3) is a local maximum.

fxx(3,-4) = -72, fyy(3,-4) = -72, so the critical point (3,-4) is a local maximum

So the extreme value of the function is a local maximum, and it occurs at the points (-4, 3) and (3, -4).

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