I want to find the global maximum and minimum of the function $f(x,y,z)=xy(1-x^2-y^2-z^2)$ on the closed unit ball? I guess they are (Maxima:$(\frac{1}{2},\frac{1}{2},0), (-\frac{1}{2},-\frac{1}{2},0)$) and (Minima:$(\frac{1}{2},-\frac{1}{2},0), (-\frac{1}{2},\frac{1}{2},0)$) but I need to prove this.
[Math] Global maximum and minimum on closed ball
calculusoptimizationreal-analysis
Best Answer
Note that maximum and minimum values only differ in sign because you can change the sign of $xy$ without violating the constraint. So it suffices to find global maxima over part of the unit ball that is in the positive orthant. Using AM-GM inequality we have $$4f(x,y,z)\leq \left(\frac{1-x^2+2xy-y^2-z^2}{2}\right)^2\\ =\left(\frac{1-(x-y)^2-z^2}{2}\right)^2\\ \leq\frac{1}{4}.$$ The first inequality becomes equations only when $2xy=1-x^2-y^2-z^2$ and the second inequality when $x=y$ and $z=0$. Thus maximum of $f$ is attained at $x=y=\frac{1}{2}$ and $z=0$. The other maxima is in the negative orthant that we excluded. Using the sign argument above minima are also at the points you mentioned.