Find the global maxima and minima

Question

Find the global maxima and minima for the function x^2+2y^2 on the interior of the triangle with the vertices (-1,2), (-1,-1), (2,-1).

Answer 1

Tom is correct except the maximum should be 9 at (-1,2) as (-1)^2 = 1

Answer 2

The objective function is x^2 + 2y^2. This is a strictly convex function.

(0,0) is in the interior and that is the global minimum of the objective function, so this is the minimum.

(To show it is in the interior: with vertices (-1, -1), (-1, 2), (2, -1), we have line segments x = -1, y = -1, and x+y = 1. (0, 0) is above y = -1, to the right of x = -1, and to the left and below x+y = 1.)

As we have a strictly convex function in a convex polygon, the maximum is at a vertex.

Thus, we just have to calculate the value of the objective function at the vertices.

For (-1, 2), (-1)^2 +2* 2^2 = 7

For (-1, -1), (-1)^2 +2* (-1)^2 = 3

For (2, -1), (2)^2 +2* (-1)^2 = 6

The maximum is 7 at (-1, 2)