Find the Maximum and Minimum of the Given Function on the Given Plane Region I've been good with most of the max/min finding in different regions, but this one's really messing with me. Can anyone lend a hand? Thanks.
$$z = 2xy$$
Region is the circular disk $x^2 + y^2 \leq 1 $
 A: Max: $ 2 \frac{\sqrt{2}}{2}\frac{\sqrt{2}}{2} = 1$.
Min: Negative of the max, by symmetry, hence $-1$.  
Hint: Do you know Lagrange multipliers? At an extremum, the gradient of $f(x, y)$ (in your case, that's $(2y, 2x)$) must be a multiple of the gradient of the constraint; in your the constraint function is $g(x, y) = x^2 + y^2 - 1$ so its gradient is $(2x, 2y)$. Hence you need 
$$
2y = c 2x \\
2x = c 2y
$$
from which you can conclude that $c = 0, 1,$ or $-1$. Then it's all downhill. 
You could have an extremum in the interior, where $\nabla f(x, y) = 0$, but in this case that happens only at the origin, and $0$ is neither a max nor min for this function. (Check the values at the points of the circle at angles $\pm \pi/4$.)
A: Hint.
Do it in polar coordinates
$$
\max_{\rho,\theta}2\rho^2\sin\theta\cos\theta, \ \ \text{s. t.}\ \ \ \rho^2\le 1
$$
now as $z$ has a saddle point at the origin as only characteristic point, the maxima/minima are located at the boundary, or at $\rho=1$ so the problem reduces to
$$
\max_{\theta}2\sin\theta\cos\theta
$$
