Question

The sum of the surface areas of a sphere and a cube is given. Show that when the sum of their volumes is least, the diameter of the sphere is equal to the edge of the cube.

 

Answer 1

Let r be the radius of the sphere, x be the side of the cube and S be the sum of the surface area of both. Then, \[S = 4\pi r^2 + 6 x^2\]

\[\Rightarrow\]\[x = \left( \frac{S - 4\pi r^2}{6} \right)^\frac{1}{2}\]     ................(1)

Sum of volumes, V= \[\frac{4}{3}\pi r^3 + x^3\]

\[\Rightarrow\] V = \[\frac{4\pi r^3}{3} + \left[ \frac{\left( S - 4\pi r^2 \right)}{6} \right]^\frac{3}{2}\]  [From eq. (1)]

\[\Rightarrow \frac{dV}{dr} = 4\pi r^2 - 2\pi r \left[ \frac{\left( S - 4\pi r^2 \right)}{6} \right]^\frac{1}{2}\]

For the minimum or maximum values of V, we must have \[\frac{dV}{dr} = 0\]       ..............(2)

\[\Rightarrow 4\pi r^2 - 2\pi r \left[ \frac{\left( S - 4\pi r^2 \right)}{6} \right]^\frac{1}{2} = 0 \left[ \text { From eq } . \left( 2 \right) \right]\]

\[ \Rightarrow 4\pi r^2 = 2\pi r \left[ \frac{\left( S - 4\pi r^2 \right)}{6} \right]^\frac{1}{2} \]

\[ \Rightarrow 4\pi r^2 = 2\pi r x ..............\left[ \text { From eq }. \left( 1 \right) \right] \]

\[ \Rightarrow x = 2r\]

Now,

\[\frac{d^2 V}{d r^2} = 8\pi r - 2\pi \left[ \frac{\left( S - 4\pi r^2 \right)}{6} \right]^\frac{1}{2} - \frac{2\pi r}{2} \left[ \frac{\left( S - 4\pi r^2 \right)}{6} \right]^{- \frac{1}{2}} \frac{\left( - 8\pi r \right)}{6}\]

\[ \Rightarrow \frac{d^2 V}{d r^2} = 8\pi r - 2\pi \left[ \frac{\left( S - 4\pi r^2 \right)}{6} \right]^\frac{1}{2} + \frac{4}{3} \pi^2 r^2 \left[ \frac{6}{\left( S - 4\pi r^2 \right)} \right]^\frac{1}{2} \]

\[ \Rightarrow \frac{d^2 V}{d r^2} = 8\pi r - 2\pi x + \frac{4}{3} \pi^2 r^2 \frac{1}{x} = 8\pi r - 4\pi r + \frac{2}{3} \pi^2 r\]

\[ \Rightarrow \frac{d^2 V}{d r^2} = 4\pi r + \frac{2}{3} \pi^2 r > 0\]

So, volume is minimum when x = 2r.