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

  • $\begingroup$ The equation of the ellipsoid is irrelevant here. $\endgroup$ – amd Jul 11 '18 at 19:46
  • $\begingroup$ Observe that both the formula for the volume and the constraint on the semi-axis lengths are symmetric in the three variables. This should immediately throw a big flag in your mind that something interesting happens when they’re equal. $\endgroup$ – amd Jul 11 '18 at 19:58
  • $\begingroup$ @Spudnick Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… $\endgroup$ – user Aug 8 '18 at 23:02

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.


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.

  • $\begingroup$ Why clutter this up by including the equation of the sphere in the Lagrangian? It adds no relevant constraints. $\endgroup$ – amd Jul 11 '18 at 19:48
  • $\begingroup$ @amd Clutter? Why not include the ellipsoid restriction? Are the results unsatisfactory? As you can observe the results are completely coherent. $\endgroup$ – Cesareo Jul 11 '18 at 20:45
  • $\begingroup$ Yes, clutter. The equation of the ellipse provides no additional useful constraints on the half-axis lengths and adds unnecessary work. $\endgroup$ – amd Jul 11 '18 at 21:57
  • $\begingroup$ In mathematics there is not the clutter concept. This is a subjective point of view. In mathematics there exist what is right and what is wrong. $\endgroup$ – Cesareo Jul 11 '18 at 23:54
  • $\begingroup$ Elegance and parsimony are important qualities in mathematical work, as is being able to recognize what is and is not relevant to a problem. The equation of the ellipsoid is irrelevant—it provides no information about the values of $a$, $b$ and $c$. $\endgroup$ – amd Jul 12 '18 at 1:23

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