finding box of maximum surface area on sphere with lagrange multipliers

Question

I want to know the largest surface area of a box with corners on the sphere with radius 4.

I tried: the surface area of one side will be $2x\cdot 2y$, so using the lagrange multiplier, the stationary points of $\Lambda = 8(xy+yz+xz) - \lambda(x^2+y^2+z^2-16)$ need to be found I think.

The partial derivatives give me the equations $8(y+z)=2\lambda x$, $8(x+z)=2\lambda y$ and $8(y+x)=2\lambda z$.

I would think that by symmetry the answer should be when $x=y=z$ so $x^2+x^2+x^2=16$, and therefore $x=\frac43 \sqrt3$.

Normally I can just substitute things and solve the system, but here I don't seem to get anywhere..

Answer 1

subtracting $8(y+z)=2λx$ from $8(x+z)=2λy$ gives $8(x-y)=2\lambda (y-x)$. so $x=y$ or $\lambda = -4$. Doing the same for another pair of equations gives us that $x=y=z$, since we don't care about solutions for lambda.

Answer 2

It is sufficient to consider axis-aligned boxes. These have a vertex ${\bf p}=(x,y,z)$ in the closed first octant, and their surface is given by $8(xy+yz+zx)$. If the box of maximal area has ${\bf p}$ in the interior of the first octant then Lagrange's method brings it to the fore. Adding the the three equations $8(y+z)=2\lambda x$, etc., you obtained one gets $16(x+y+z)=2\lambda(x+y+z)$, and as $x+y+z>0$ in the interior of the first octant we conclude that $\lambda=8$, which then leads to $x=y=z={4\over\sqrt{3}}$. The area in question then would be $128$.

But we also have to consider points ${\bf p}=(x,y,z)$ on one of the coordinate planes. It is sufficient to consider the plane $z=0$, so that we have to maximize $8xy$ under the constraints $x\geq0$, $y\geq0$, $x^2+y^2=16$. It is well known that this maximum occurs at $x=y=\sqrt{8}$, so that the surface of this degenerate box becomes $64$ only.

To sum it up, we have proven that the rectangular box of maximal surface area inscribed in a sphere of radius $4$ has surface area $128$.