Maximum und Minimum I need help for c) also I solved a and b, but c) I could not.
Let $f~:~\Bbb R^2\to \Bbb R$ be defined as $f(x,y)=e^{-x^2-y^2+2x-2y}$
a) Determine all local extrema for $f$.
b) Determine all potential extrema of $f$ under the condition $x^2+y^2=4$.  Decide whether each is a maximum or minimum.
c) Let $D$ be the region defined as $D=\{(x,y)\in\Bbb R^2~|~x^2+y^2\leq 4\}$.  Determine the maximum and minimum of $f$ over the region $D$.
Thanks
 A: Hints: This is basically a question on the optimization of a function in multiple variables. 
For a), the gradiant of $f$ must be zero, that is $$\nabla f= \begin{pmatrix}\frac{\partial f}{\partial x} \\\frac{\partial f}{\partial y}\end{pmatrix}$$
must be $\vec{0}$. For the other parts use Lagrange multipliers. You have the constraint $$g(x,y) = x^2+y^2-4=0$$
And you must solve the system of equations
$$\nabla f = \lambda \nabla g$$
Together with the constraint $x^2+y^2=4$, of course. That's a system of equations in 3 equations and 3 variables ($x,y,\lambda$). 
You will then receive candidates for $(x,y)$. You can check with the Hesse-matrix wether these points are minima, maxima or saddle-points. Hope this gives you a starting point. Once you have the $(x,y)$-values of the minima/maxima, part c) is trivial, since you just need to compute the function value of these points. You see that the upper constraint was the boundary of the circle of radius $2$, that is $x^2+y^2=4$, now the region $D$ is that boundary and everything inside. The maxima and minima is either at the boundary (which was explored in exercise (b)) or in the general inside of it (which is solved in exercise (a)). You can conclude the minima and maximima inside $D$ now.
A: The global minimum is either in the interior or at the boundary. If they are at the boundary, you have solved them in part (b). If they are in the interior, you have solved them in part (a).
Just compare the solutions at part (b) and also solutions from part (a) that are inside the domain $D$.
A: Part (c) asks you to find maximum and minimum values of $e^{-x^2- y^2+ 2x- 2y}$ restricted to the disk $x^2+ y^2\le 1$.  There is a theorem that says that a continuous function takes on both maximum and minimum values on a closed and bounded set.  Further, such maximum or minimum values occur either in the interior where the partial derivatives are 0 or on the boundary.  You apparently have already determined where the partial derivatives are 0.  Are they inside this circle?  If so what are the values of the function there? 
  On the circular boundary, $x^2+ y^2= 1$, the function is $e^{-1+ 2x- 2y}= e^{2x}e^{-2y}/e$.  I think I would be inclined to convert to polar coordinates: x= cos(t), y= sin(t) so the function is $e^{2cos(t)}e^{-2sin(t)}/e$.  Where is the derivative with respect to t equal to 0?  If there are any such places, evaluate the function there.  You might also need to evaluate at $t= 0$ and $t= 2\pi$.  The maximum and minimum values of the function on the disk will be the largest and smallest of those values.
