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

Question

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 \ \le \ 4 \ $.

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

Answer 1

You will probably find the solution much more obvious if you convert it to a function of $r$ and $\theta$.

Answer 2

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 $r_0^2e^{-r_0^2}(1+ \sin^2 \theta_0).$

Answer 3

As has already been pointed out, the function $ \ f(x,y) \ = \ (x^2+2y^2)·e^{-x^2-y^2} \ = \ (x^2+2y^2)·e^{-(x^2+y^2)} \ $ is non-negative; since the exponential ("decaying") factor is always positive, this function is only equal to zero at the origin, so that is the location of the absolute minimum.

What we may also say about this function is that it has "four-fold" symmetry $ \ [ \ f(-x,y) \ = \ f(x,-y) \ = \ f(-x,-y) \ = \ f(x,y) \ ] \ $ , so that any critical points not on a coordinate axis will appear in "sets of four" and those on an axis will form pairs. If the function were simply $ \ (x^2+y^2)·e^{-(x^2+y^2)} \ , \ $ the function would have "radial symmetry", and we know from a frequently-posed problem that the maximum of $ \ r^2·e^{-r^2} \ $ occurs at radial distance $ \ r \ = \ 1 \ $ , thus at all points on a unit circle centered at the origin. For our function, the factor $ \ (x^2+2y^2) \ $ has the effect of "stretching" the radial function "away from" the $ \ y-$ axis, turning the circle into an ellipse with its major axis on the $ \ x-$ axis. This "breaks" the radial symmetry down to the four-fold symmetry of $ \ f(x) \ $ . Since the distance-squared $ \ (x^2+y^2) \ $ is increased for points on or outside the unit circle but not on the $ \ y-$ axis, the exponential factor decreases "more rapidly" than $ \ (x^2+2y^2) \ $ increases, so we might intuit that $ \ f(x) \ $ is only maximal at the points where the unit circle intercepts the $ \ y-$ axis.

We can check this by exploiting the four-fold symmetry and considering "slices" through the function on the lines $ \ y = \pm \ mx \ $ to produce $ \ \phi(m,x) \ = \ [ \ (1+2m^2)x^2 \ ]·e^{-[ \ (1+m^2)x^2 \ ]} \ $ . (This is a variation on the idea of using polar coordinates.) On a particular such slice, fixing $ \ m \ $ , the extremization calculation gives us

$$ [ \ 2 \ (1+2m^2)x \ ]·e^{-[ \ (1+m^2)x^2 \ ]} \ - \ [ \ 2 \ (1+m^2)x \ ]·[ \ (1+2m^2)x^2 \ ]·e^{-[ \ (1+m^2)x^2 \ ]} \ \ = \ \ 0 $$

$$ \Rightarrow \ \ \ x \ \ = \ \ (1+m^2) \ x^3 \ \ \ \Rightarrow \ \ \ x \ = \ 0 \ \ , \ \ x^2 \ = \ \frac{1}{1 \ + \ m^2} \ \ . $$

We already know about $ \ x \ = \ 0 \ $ ; inserting our second result into $ \ \phi(m,x) \ $ tells us that the maximum value on the slice is

$$ \left( \frac{1+2m^2}{1+m^2} \right)·e^{- \left( \frac{1+m^2}{1+m^2} \right)} \ \ = \ \ \left( \frac{1+2m^2}{1+m^2} \right)· \frac{1}{e} \ \ . $$

For the (two) critical points on the $ \ x-$ axis ( $ m = 0 \ \ \Rightarrow \ \ x \ = \ \pm 1 $ ) , the maximum of the function is $ \ \frac{1}{e} \ $ . We see that $$ \ \lim_{m \ \rightarrow \ \infty} \ \left( \frac{1+2m^2}{1+m^2} \right)· \frac{1}{e} \ \ = \ \ \frac{2}{e} \ \ , $$

so our expectation that the maxima of $ \ f(x,y) \ $ lie on the $ \ y-$ axis ( $ \lim_{m \ \rightarrow \ \infty} \ \frac{1}{1 \ + \ m^2} \ = \ 0 $ ) was correct. The absolute maxima are confirmed as $ \ f(0,1) \ = \ f(0,-1) \ = \ \frac{2}{e} \ \ $ .