# Determine the rectangular parallelepiped of maximum surface area which can be inscribed in a sphere.

The equation of a sphere of radius $$r$$ centered at the origin $$(0,0,0)$$ is given by $$x^2 + y^2 + z^2 = r^2$$. The surface area of a rectangular parallelepiped with center at the origin is given by $$S = 8(xy+yz+zx)= 8\left[xy +(x+y)\sqrt{r^2 - x^2 - y^2}\right].$$ For the maximum surface area, we have the necessary conditions $$\frac{\partial S}{\partial x}=0$$ and $$\frac{\partial S}{\partial y}=0$$. Therefore, differentiating $$S$$ with respect to $$x$$ and $$y$$, we have $$\frac{\partial S}{\partial x} = 8\left[y+\sqrt{r^2 - x^2 - y^2}- \frac{x(x+y)}{\sqrt{r^2 - x^2 - y^2}}\right]$$ and $$\frac{\partial S}{\partial y} = 8\left[x+\sqrt{r^2 - x^2 - y^2}- \frac{y(x+y)}{\sqrt{r^2 - x^2 - y^2}}\right].$$

To find critical points, we equate the derivatives to zero. As $$\sqrt{r^2 - x^2 - y^2} \ne 0$$ in a sphere, we get $$y\sqrt{r^2 - x^2 - y^2}+r^2 - x^2 - y^2- x(x+y) = 0$$ and $$x\sqrt{r^2 - x^2 - y^2}+r^2 - x^2 - y^2- y(x+y) = 0.$$

Subtracting, we get $$(y-x)\left[\sqrt{r^2 - x^2 - y^2}+(x+y)\right]=0$$ $$\Rightarrow$$ $$y=x$$ or $$\sqrt{r^2 - x^2 - y^2}+(x+y)=0$$.

Then how to find the relation between $$x$$, $$y$$ and $$z$$.

• Would you prefer not to use Lagrange multipliers? Commented Feb 27 at 10:39

The constraint is $$x^2+y^2+z^2=r^2.$$ The surface area of a rectangular parallelepiped with center at the origin, axes perpendicular to the $$x$$, $$y$$, and $$z$$ axes, and one vertex at the point $$(x,y,z)$$ ($$x,y,z\geqslant 0$$) is $$S=8(xy+yz+zx).$$ I'm going to look at the objective function $$T=xy+yz+zx.$$

We treat $$x,y$$ as the independent variables, and $$z$$ is a function of them, as determined by the constraint. Note that $$z^2=r^2-x^2-y^2,$$ so $$2z\frac{\partial z}{\partial x} = -2x,$$ and $$\frac{\partial z}{\partial x} = -\frac{x}{z}.$$ By symmetry, $$\frac{\partial z}{\partial y}=-\frac{y}{z}.$$

In principle, we should be worried about when $$z=0$$, but this has surface area zero and will be a minimum, not a maximum, surface area.

Then $$\frac{\partial T}{\partial x} = y+y\frac{\partial z}{\partial x}+x \frac{\partial z}{\partial x}+z$$ and $$\frac{\partial T}{\partial y} = x+x \frac{\partial z}{\partial x}+y \frac{\partial z}{\partial y}+z.$$

Using $$\frac{\partial z}{\partial x}=0,$$ we get $$yz+z^2=xy+x^2\tag{1}.$$ Using $$\frac{\partial z}{\partial y}=0$$, we get $$xz+z^2=xy+y^2\tag{2}.$$

Subtracting $$(2)$$ from $$(1)$$ gives $$(y-x)z=x^2-y^2=(x-y)(x+y).$$ Either $$x=y$$, or we can divide through by $$x-y$$ to get $$-z=x+y$$. But the latter implies $$x=y=z$$, which obviously doesn't maximize the surface area. So it must be the case that $$x=y$$ at our critical point.

Going back to $$(1)$$ and plugging in $$x=y$$, we get $$x^2+xz-2x^2=0,$$ and we can factor this as $$(z+2x)(x-z)=0.$$ Since $$z+2x=0$$ implies $$z=x=0$$, which doesn't maximize surface area, we get $$x=z$$. Thus we have $$x=y=z$$ at our surface area-maximizing critical point.

• I thank you for providing the answer Commented Feb 27 at 11:47