Find absolute extrema

Question

Find the absolute extrema of the function f(x,y) = (3x−y)². over the triangular region in the xy-plane with vertices (2,0), (0,2), and (1,3).

a. absolute maximum: 36 at (2,0)

absolute minimum: 0 at (1.3) and along the line y = 3x

b. absolute maximum: 36 at (2,0) absolute minimum: 1 at (1,2)

C. absolute maximum: 1 at (1,2) absolute minimum: 0 at (1,3) and along the line y = 3x

d. absolute maximum: 36 at (2,0)

absolute minimum: 4 at (0,2)

e. absolute maximum: 4 at (0,2)

absolute minimum: 0 at (1.3) and along the line y = = 3x

Answer 1

You don't have to use calculus for this problem. (3x-y)^2 = 0 on y = 3x, and this is the minimum. The function is constant on any line with slope 3 and is otherwise strictly convex along any line. Thus, the maxima will be at the vertices (the slope does not equal 3 on any boundaries). Thus, plugging in the three vertices, the maximum is at (2, 0) and is 36.

The answer is a, though no reason to specifically mention (1, 3) unless we also want to point out the other endpoint of the line segment, which is (1/2, 3/2) - this is where y = 3x meets x + y = 2, one of the boundaries.