<s>[INST] ellipsoid of greatest volume is a sphere The equation of an ellipsoid is $$f(x,y,x)=(\frac xa)^2+(\frac yb)^2+(\frac zc)^2=1$$.
Given that the volume of an ellipsoid is $$V=\frac43\pi abc$$ and the constraint $$L=a+b+c$$ L some positive constant. Show that the ellipsoid with greatest volume is a sphere.
I should use Lagrange multipliers for this question.
I tried doing $$\nabla f = \lambda\nabla V$$ which gave an anwser making no sense$$<\frac{2x}{a^2},\frac{2y}{b^2},\frac{2z}{c^2}> =\lambda <0,0,0>$$
So then I tried $$"\nabla"V=\lambda"\nabla"L$$
where $"\nabla"$ treats a as x, b as y, c as z, and got
$$4\pi/3<bc,ac,ab>=\lambda<1,1,1>$$
so $ab=ac=bc$ gives $a=b=c$ a sphere. But why am  I allowed to use $"\nabla"$ as such
 [/INST] Here the lagrangian reads
$$
L(a,b,c,x,y,z,\lambda,\mu) = \frac 43 \pi a b c +\lambda\left(\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2+\left(\frac{z}{c}\right)^2-1\right)+\mu(a+b+c-L_0)
$$
The stationary conditions give
$$
\nabla L = \left\{
\begin{array}{rcl}
 -\frac{2 \lambda  x^2}{a^3}+\mu +\frac{4 b c \pi }{3}=0 \\
 -\frac{2 \lambda  y^2}{b^3}+\mu +\frac{4 a c \pi }{3}=0 \\
 -\frac{2 \lambda  z^2}{c^3}+\mu +\frac{4 a b \pi }{3}=0 \\
 \frac{2 \lambda  x}{a^2}=0 \\
 \frac{2 \lambda  y}{b^2}=0 \\
 \frac{2 \lambda  z}{c^2}=0 \\
 \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-1=0 \\
 a+b+c-L_0=0 \\
\end{array}
\right.
$$
and solving gives
$$
\left[
\begin{array}{ccccccccc}
a & b & c & x & y & z & \lambda & \mu & V\\
 \frac{L_0}{3} & \frac{L_0}{3} & \frac{L_0}{3} & x & y & -\frac{1}{3} \sqrt{L_0^2-9 x^2-9 y^2} & 0 & -\frac{4 L_0^2
   \pi }{27} & \frac{4 L_0^3 \pi }{81} \\
 \frac{L_0}{3} & \frac{L_0}{3} & \frac{L_0}{3} & x & y & \frac{1}{3} \sqrt{L_0^2-9 x^2-9 y^2} & 0 & -\frac{4 L_0^2 \pi
   }{27} & \frac{4 L_0^3 \pi }{81} \\
\end{array}
\right]
$$
As can be observed the volume is maximum for $a = b = c\;$ defining a sphere. Note also the found relationship
$$
9z^2 = L_0^2-9x^2-9y^2
$$
which is a sphere.
</s>