Find maximum and minimum of $f(x,y) = x^2 y^2 – 2x – 2y$ in $0 \leq x \leq y \leq 5$.
So first we need to check inside the domain, I got only one point $A(1,1)$ where $f(1,1) = -3$. and after further checking it is a saddle point.
Now we want to check on the edges, we have 3 edges: $(1) x=y, 0\leq x\leq y \leq 5$ and $(2) y=5, 0 \leq x \leq 5$ and $(3) x=0, 0\leq y \leq 5$.
So I started with each of them, using Lagrange. I started with the first and got:
$l(x,y,\lambda) = x^2y^2 – 2x – 2y + \lambda (x-y)$.
$l_x(x,y) = 2xy^2 – 2 + \lambda = 0$
$l_y(x,y) = 2x^2y – 2 – \lambda = 0 $
$x-y = 0 \rightarrow x=y$.
But then what to do ? I always get $x,y = \sqrt[3]{1+ 0.5\lambda}$, which gets me nowhere. Any help would be appreciated
Best Answer
Well, we have $f(x,y)=x^2y^2-2x-2y$ considered on the following green area:
So $f_x=2xy^2-2,~~f_y=2x^2y-2$ and solving $f_x=f_y=0$ gives us the critical point $x=y=1$ which is on the border of the area. Now think about the border: $$y=5,~~ 0\le x\le 5 \;\;\;\; x=0,~~0\le y\le 5 \;\;\;\; y=x,~~ 0\le x\le 5$$
If $y=5,~~ 0\le x\le 5$, so $$f(x,y)\to f(x)=25x^2-2x-10$$ and this easy for you to find the critical points which are maybe the best candidates on this border.
If $x=0,~~0\le y\le 5$, so $$f(x,y)\to f(y)=-2y$$ there is a similar way of finding such points on this border.