# Finding the absolute maximum and minimum of a function in a disk?

Find the absolute maximum and minimum of the function $e^{-x^2-y^2}(x^2+2y^2)$ in the disk $x^2 + y^2 = 4$.

I found the partial derivatives but am unable to find the critical points and proceed further

• Please check that the edit I made is what you meant (since it was ambiguous). – Arnaud Mortier May 23 '18 at 22:56
• Could you also include the partial derivatives that you found? – Arnaud Mortier May 23 '18 at 22:56
• I used polar coordinates to transform the function to $f(r,\theta)=r^2e^{-r^2}(1+\sin^2\theta)$. Then I took partial derivatives and considered the boundary $r=2$ to get 16 critical points $\{0,1,2\}\times\{0,\pi/2,\pi,3\pi/2\}$. The minimum was $0$ whenever $r=0$ and maximum was $2/e$ at $(1,\pi/2)$ and $(1,3\pi/2)$. – M. Nestor May 23 '18 at 23:14

You will probably find the solution much more obvious if you convert it to a function of $r$ and $\theta$.
This problem is much easier if you avoid partial derivatives entirely. The minimum announces itself right away, as this function is $\ge 0$ and equals $0$ at $(0,0).$ The minimum is therefore $0.$
For the maximum, go to polar coordinates to get $r^2e^{-r^2}(1+ \sin^2 \theta).$ Find the $r_0\in [0,2]$ that maximizes $r^2e^{-r^2},$ then find the $\theta_0 \in [0,2\pi]$ that maximizes $1+ \sin^2 \theta.$ For the first function, this is an easy one variable calculus problem. For the second function, the maximum is clear. The maximum value then has to be $r_0^2e^{-r_0^2}(1+ \sin^2 \theta_0).$