absolute extrema of f(x,y) in a closed triangle

Question

Find the absolute extrema of f(x,y) = x2y − 2x + y on the closed triangle T with vertices (1,0),(5,0) and (1,4).

Answer 1

The partial derivative of f with respect to y is x² +1 . This expression can never be zero. Thus f can have neither an extreme max, nor an extreme min in the interior of the triangle.

This means that the boundaries must be checked. There are three of them: a vertical line segment, a horizontal line segment and a line with the equation y = 5 -x.

It is easy to check that f does not have a max or min on the interior of either of the first two line segments.

Using the substitution y = 5-1 in f, a TI-84 calculator can be used to find that f has a maximum of f =14 at the point (3,2). The values of f at the three corners must also be checked

f(1,0) = -2 f(5,0) = -10 and f(1, 4) = 6

Studying the points at the corners and the point (3,2) , we see that the absolute max is f =14 at (3,2) and an absolute min of f = -10 at (5,0)