Absolute Maximum and Minimum Question

Question

Find the greatest and the least values of the function $z=x^3+y^3-3xy$ on the region ${(x,y): 0≤x≤2, -2≤y≤2}$.

I tried solving $∇f=0$, but the answer provided makes nk sense. Please help.

Answer 1

We are told to find the global extrema of the function $f(x,y):=x^3+y^3-3xy$ on the rectangle $R:=[0,2]\times[{-2},2]$. This $R$ is a stratified set, having an open interior, the relative interiors of the four edges, and the four vertices. If the maximum of $f$ on $R$ is taken in the interior it is a zero of the gradient $\nabla f$, if it is taken on one of the edges it is a zero of $\psi'$, where $\psi$ is the restriction of $f$ to this edge. We therefore have to set up a candidate list consisting of the mentioned special points and the four vertices.

You already have determined the zero $(1,1)$ of $\nabla f$ in the interior of $R$. On the western edge of $R$ we have to look at $\psi_{\rm w}(y):=f(0,y)=y^3$ $(-2<y<2)$ and find there the conditionally stationary point $(0,0)$. On the southern edge of $R$ we have to look at $\psi_{\rm s}(x):=f(x,-2)=x^3-8+6x$ $(0<x<2)$. Since $\psi_{\rm s}'(x)=3x^2+6>0$ there are no conditionally stationary on this edge. On the eastern edge we have to look at $\psi_{\rm e}(y):=f(2,y)=8+y^3-6y$ with $\psi_{\rm e}'(y)=3y^2-6$. This leads to the conditionally stationary points $(2,\pm\sqrt{2})$. Finally on the northern edge of $R$ we have to look at $\psi_{\rm n}(x):=f(x,2)==x^3+8-6x$ $(0<x<2)$ with $\psi_{\rm n}'(x)=3x^2-6$, leading to the conditionally stationary point $(\sqrt{2},2)$.

The candidtate list $L$ is therefore given by $$L=\bigl\{(1,1),(0,0),(2,\sqrt{2}),(2,-\sqrt{2}),(\sqrt{2},2),(0,-2),(2,-2),(2,2),(0,2)\bigr\}\ .$$ You now have to compute and compare the values of $f$ in the points of the finite set $L$ in order to find the global extrema of $f$ on $R$.

Answer 2

You forgot to "look" on the boundary. If you consider the derivatives, you implicitly assume that the minimum or maximum is in $]0,2[\times]-2,2[$ since on the boundary, the derivative is not defined. So you still need to parametrize the boundary by a curve (4 different lines) and then compare the results.

If you do not know the result already, you have to do this for any of the four lines that constitute the boundary. If we do it for the line $x=2$, $y \in [-2,2]$, we want to maximize $2^3 + y^3 - 3*2*y$. Taking derivatives, we see that $y=\sqrt{2}$ or $y=-\sqrt{2}$ are extrema. If we take $y=-\sqrt{2}$, we get as a result $8 + 4\sqrt{2}$