Find the absolute maximum and minimum values of the function f on the set d. f (x, y) = 4xy 2 − x 2 y 2 − xy 3 ; d is the closed triangular region in the xy-plane with vertices (0, 0), (0, 6), and (6, 0).

For now, ignore any critical points that occur on the boundary of .

Case 1: If , then either or , so we have two critical points (0, 0) and (4, 0), both on the boundary.

Case 2a: ... if , then - critical point at (0, 4) on the boundary.

Case 2b: ... if , then - critical point at (2, 0) on the boundary.

Case 2c: ... if , then

- critical point at (1, 2) not on the boundary.

We compute the Hessian matrix of :

which means (1, 2) is a local maximum, where we get .

Now we consider the boundary in three parts:

(1) Fix . Then . It's possible that 0 is the smallest value of whereever you go.

(2) Fix . Then , and we get the same result as in the previous case.

(3) Fix . Then . Denote this function by . We compute the first derivative and find the critical points:



then compute the second derivative and determine its sign at these points:



which indicates that a maximum occurs as and a minimum at . Respectively, we have and , and and .

So the absolute maximum of over is 4 at (1, 2), and the absolute minimum is -64 at (2, 4).

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step-by-step explanation: