Maximum of a polynomial function on the unit sphere in $\mathbf R^3$

Question

Given three positive real numbers $a,b,c>0$, consider the following function, defined on $\mathbf R^3$ $$ f(x,y,z) := y^2(a^2x^2+b^2y^2+c^2z^2). $$

Question. What is the maximum of $f$ on $\mathbf S^2 :=\{(x,y,z): x^2+y^2+z^2=1\}$ (in terms of $a,b,c$)?

The existence of the maximum on $\mathbf S^2$ is a trivial consequence of Weierstrass' theorem. Lagrange multipliers haven't been very illuminating, and after some tedious computations I gave up. Spherical coordinates? Seems a mess, and nothing simplifies. What bothers me the most is that I suspect there is a nice geometric interpretation of all of this, but somehow I fail to see it. Any suggestions?

Answer 1

Assuming wlog $a\ge c$, the maximum on $f$ on $\mathbf S^2$ equals the maximum on $\mathbf S^1$ of $y^2(a^2x^2+b^2y^2)$, i.e. (letting $t:=y^2$) the maximum of $$g(t):=a^2t+(b^2-a^2)t^2\text{ on }[0,1].$$

• If $a=b$, this maximum is obviously $a^2$.
• If $a\ne b$ then $g'(t)=a^2+2(b^2-a^2)t$ equals $0$ iff $t=t_0:=\frac{a^2}{2(a^2-b^2)}$.
  • If $t_0\in[0,1]$, i.e. if $b\le\frac a{\sqrt2}$, this maximum is $g(t_0)=\frac{a^4}{4(a^2-b^2)}$.
  • If $b\ge\frac a{\sqrt2}$, this maximum is $g(1)=b^2$.

Answer 2

Less geometric than algebraic, but here's an approach: notice that $x, y, z$ only ever appear in the function and constraint as their squares -- so write $u=x^2, v=y^2, w=z^2$ and reduce to the much simpler optimization of $v \cdot (a^2u + b^2v + c^2w)$ subject to $u+v+w=1$ and $u,v,w\geq0$. At this point, you could skip the Lagrange multipliers here by eliminating $w = 1 - u - v$ too.