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Let $T\subset\mathbb{R}^2$ be the (closed) triangle bounded by the lines $x+y=4$, $x\ge-1$ and $y\ge-1$. I want to find and classify all the extrema of the function $f(x,y)=-x^2y(x+y-2)$ on the triangle $T$.

I've done the following: Solving $\nabla f=0$ gives the points $(0,0), (0,1)$ and $0,2$. But when I compute the Hessian, I keep getting eigenvalue 0.

Is there someone who can show me how to solve this problem?

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But according to your function, the critical points are given as $(0,y)$ $(2,0)$ $(1,\frac{1}{2})$. Check it again. –  B. S. Jun 20 '12 at 18:03

1 Answer 1

Find the critical points of $f$. Ignore those that are not in $T$. Find the maximum and minimum of $f$ on the left side; that is, maximize and minimize $f(-1,y)$ for $-1 \leq y \leq 5$. Find the maximum and minimum of $f$ on the bottom side; that is, maximize and minimize $f(x,-1)$ for $-1 \leq x \leq 5$. Find the maximum and minimum of $f$ on the hypotenuse; that is, maximize and minimize $f(x,4-x)$ for $-1 \leq x \leq 5$. Now take the maximum and the minimum of the values of $f$ you found at the critical point(s) and for the three sides. You don't need to take the Hessian of anything; indeed, it is sometimes a waste of time. The procedure is based on the fact that a continuous function on a compact subset of Euclidean space has a (global) maximum and minimum, which occur at critical points or on the boundary.

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