Extreme values of $f(x, y)=\frac{(x+y)^{2}}{2}+\frac{(x-y)^{3}}{3}$ on $D=\left\{(x, y) \in \mathbb{R}^{2}:|y| \leq 1-|x|\right\}$

Question

Extreme values of $f(x, y)=\frac{(x+y)^{2}}{2}+\frac{(x-y)^{3}}{3}$ on $D=\left\{(x, y) \in \mathbb{R}^{2}:|y| \leq 1-|x|\right\}$

I resolved by taking the partial derivatives and computing the Hessian matrix, so $(0,0)$ is a saddle point.

For the first edge I tried to parametrize $D$ $$\begin{cases}x=1-t\\y=t\end{cases}$$ and resolve for $f(x,y)$: $$f(t) = \frac{1}{2} + \frac{(2t-1)^3}{3}$$ $$f'(t)=4t-2=0 \implies t=\frac{1}{2}$$ and follows that $\left(\frac{1}{2},\frac{1}{2}\right)$ could be and extreme.

It's clear that $(1,0)$ is a max value, but I can't find it. Can someone help me out in this?

Answer 1

Substitute $u=x+y,v=x-y$ to transform the problem into finding the extreme points of $\frac{u^2}2+\frac{v^3}3$ on $u,v\in[-1,1]$. Then it is clear that the maxima are at $u=\pm1,v=1$ and the minimum is at $u=0,v=-1$, which when translated to the $xy$ plane gives the maxima at $x=0,y=-1$ and $x=1,y=0$ with $f(x,y)=\frac56$, and the minimum at $x=-\frac12,y=\frac12$ with $f(x,y)=-\frac13$.