Finding the Absolute Maximum and Minimum of a 3D Function Find the absolute maximum and minimum values of the function:
$$f(x,y)=2x^3+2xy^2-x-y^2$$
on the unit disk $D=\{(x,y):x^2+y^2\leq 1\}$.
 A: We first take the partial derivatives with respect to each variable and set them to zero:
$$\frac{\partial f}{\partial x}=6x^2 + 2y^2 - 1=0$$
$$\frac{\partial f}{\partial y}=4xy - 2y=0$$
From the second equation, we have either: $y=0$ or $x=\frac{1}{2}$. Now we substitute each value into the first equation to get the corresponding variable: for $y=0$, we get $x=\pm \frac{1}{\sqrt{6}}$, while for $x=\frac{1}{2}$ we have no real solutions, so we discard this. 
Note that both $(0,\frac{1}{\sqrt{6}})$ and $(0,-\frac{1}{\sqrt{6}})$ satisfy our constraint. We evaluate $f(0,\frac{1}{\sqrt{6}})$ and $f(0,-\frac{1}{\sqrt{6}})$ which give us $f=-0.27$ and $f=0.27$, respectively.
We now need to check the perimeter of the unit disk. On this disk, $y^2 = 1-x^2$, so we substitute this into the original function to get:
$$ f = x^2 + x - 1 $$
We set $df/dx=0$ to get $x=-\frac{1}{2}$, and we evaluate $f$ at this point to get $-\frac{5}{4}$.
Finally we have to check the extreme values of $x$ on the unit disk: $x=-1$ and $x=1$, which give us $f=-1$ and $f=1$, respectively.
You now have your global minimum and maximum.
A: Hint: for the interior of disk you may use usual condition $\nabla f(x,y) = 0$. For the boundary of the disk you may use parametrisation $x = \cos \phi$, $y = \sin \phi$ and minimize it w.r.t. to $\phi$ (remember, it's $2\pi$-periodic, so you should investigate it only on segment $\lbrack 0, 2\pi )$ really). Or you may use Lagrange multipliers method.
