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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 $z$.

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    $\begingroup$ Would you prefer not to use Lagrange multipliers? $\endgroup$
    – user469053
    Commented Feb 27 at 10:39

1 Answer 1

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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.

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  • $\begingroup$ I thank you for providing the answer $\endgroup$ Commented Feb 27 at 11:47

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