Volume is Minimum

Let the sum of their total surface areas be $S$ , then we have:

$\displaystyle S=4 \pi r^{2} + 6x^{2}$

Differentiating with respect to $r$ ,

$\displaystyle \implies 0 = 8 \pi r + 12x(\frac{dx}{dr})$

$\displaystyle \implies \boxed{\frac{dx}{dr}=- \frac{2 \pi r}{3x}}$

Now, let the total volume of the sphere and the cuboid be $V$ , then:

$\displaystyle V = \frac{4}{3} \pi r^{3} + \frac{2}{3} x^{3}$

Differentiating with respect to $r$ ,

$\displaystyle \implies \frac{dV}{dr} = 4 \pi r^2 + 2x^{2} (\frac{dx}{dr})$ .

Clearly, $V$ is minimum when $\frac{dV}{dr} = 0$ , hence,

$\displaystyle 4 \pi r^2 + 2x^{2} (- \frac{2 \pi r}{3x}) = 0 \implies r=\frac{x}{3}=\frac{x}{\boxed{2} + 1} \implies \boxed{n=2}$