A problem asks me to find the absolute extrema of the function given by $f: \mathbb{R}^2 \rightarrow \mathbb{R} ,f(x,y)=(x^2+y^2)e^{-(x^2+y^2)}$.
Now, how can I find the critical points?.
As far as I know it should be done with the partial derivatives, but here I have that $\displaystyle\frac{\partial f}{\partial x} = 2xe^{-(x^2+y^2)}(1-x^2-y^2)$ and $\displaystyle\frac{\partial f}{\partial y}=2ey^{-(x^2+y^2)}(1-x^2-y^2)$. Then all the critical points seem to be $\{(x,y)\in \mathbb{R}^2 : x^2+y^2=1\}\cup\{(x,y):x,y\in\mathbb{R}\}$
Since $(x^2+y^2)$ and $e^{-(x^2+y^2)}$ are both greater than zero I can say that $(0,0)$ is an absolute minima (right?), but what about the maxima?.
I can't picture how every point in the circunference can be a critical point. I plotted the function and it seems that the functions increase and decrease several times, how can I test the function to find absolute maxima (or relative) around $g(x,y) = 1-x^2-y^2$ ?.
I also tried to solve this with Lagrange multipliers but implies solving the system given by
$\begin{cases} 2xe^{-(x^2+y^2)}(1-x^2-y^2) = -2\lambda x & (1) \\ 2ey^{-(x^2+y^2)}(1-x^2-y^2) = -2\lambda y & (2)\\ 1-x^2-y^2=0 & (3)\end{cases}$
And I can't find $\lambda$ because I get the same solution for $\lambda$ with $(1)$ and $(2)$, this is, $\lambda =-2e^{-(x^2+y^2)}(1-x^2-y^2)$
.
Any hints?.
Best Answer
Expanding on what other have said, if you want to avoid using partial differentiation then you need to make the substation $r^2 = x^2 + y^2$ which converts to polar coordinates. Then the differential is simplified to
$$ \frac{d}{dr} (r^2e^{-r^2}) = e^{-r^2} (2r-2 r^3) $$
Equating this to $0$ gives the roots $r = 1$ and $r = 0$, or $x^2 + y^2 = 1$ and $x^2 + y^2 = 0$, which correspond to turning points on the unit circle, and at the point $(0,0)$.
Edit
To verify the extrema you can find the second derivative of the expression above in $r$. In multi-vartiate calculus (like the original problem), you would need to find the hessian and check if it is positive definite (local minimum), or negative definite (local maximum). Note that the hessian is the multi-variate extension of the second derivative. Also, be careful that the hessian test only verifies local extrema and can not be used to show a general point is a global optimum unless the problem is convex.
The second derivative with respect to $r$ is
$$ \frac{d^2}{dr^2} (r^2e^{-r^2}) = \frac{d}{dr} (e^{-r^2} (2r-2 r^3)) = 2 e^{-r^2} (1-5 r^2+2 r^4) $$
When $r = 0$ we find the second derivative to be $2$, thus the point $(0,0)$ is a local minimum. When $r = 1$ we find the second derivative to be $-4/e$, thus the points on the unit circle are local maximums.
Now in terms of showing the points are in-fact global extrema, you can appeal to the nature of the problem. It is not difficult to show that outside the unit circle the function is always decreasing, but is bounded by the positive axis.