Find the maximum and minimum of a function with three variables

Question

The problem is finding the maximum and minimum of the following three variables function:

$f(x, y, z)=(2x^2+2y^2+z^2+2yz+2zx)e^{-(x^2+y^2+z^2)}$, $(x, y, z)\in \Bbb R^3$

I noticed since

$f(x, y, z)=(2x^2+2y^2+z^2+2yz+2zx)e^{-(x^2+y^2+z^2)}=[(\sqrt 2x +\frac{\sqrt 2}{2}z)^2+(\sqrt 2y +\frac{\sqrt 2}{2}z)^2]e^{-(x^2+y^2+z^2)}\ge0$

and the equality holds iff

$(x, y, z)=t(2, 2, -1)$, $t\in \Bbb R$,

Thus the minimum of $f$ is $0$. But now I don't know how to get the maximum. There is no way to apply Lagrange multiplier since there's no contraints. I also thought of using Hessian matrix, but there's just too much calculation and I don't think that's the best way to do it. Any help would be appreciated!

Answer 1

Since $\lim_{\|(x,y,z)\|\to\infty}f(x,y,z)=0$, the function $f$ has to have a maximum at some point, and then $\nabla f$ will have to be $(0,0,0)$ at that point. If you solve the system $\nabla f(x,y,z)=(0,0,0)$, you will recover the points that you already have (those at which $f$ has a minimum), and you will also get:

• $\pm\left(\frac1{\sqrt2},-\frac1{\sqrt2},0\right)$, at which $f$ take the value $\frac 2e$;
• $\pm\left(\frac1{\sqrt3},\frac1{\sqrt3},\frac1{\sqrt3}\right)$, at which $f$ take the value $\frac 3e$.

Therefore, the maximum is $\frac3e$.