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
Best Answer
We are applying the Lagrange multiplier method to $V=V(a,b,c)$ as a function of the three parameters $a,b,c$ with the constraint $a+b+c=L$ and therefore the gradient needs to be evaluated with respect to those variables.