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 us assume radius of sphere be ‘r’ and length of side of cube is ‘l’

We know that, 

⇒ Surface area of sphere = 4πr² 

⇒ Surface area of cube = 6l² 

According the problem, 

the sum of surface areas of a sphere and cube is known. Let us assume the sum be S 

⇒ S = 4πr² + 6l² …… (1) 

We also know that, 

⇒ Volume of sphere = \(\frac{4}{3}\)πr³ 

⇒ Volume of cube = l³ 

We need the sum of volumes to be least. 

Let us assume the sum of volumes be V

⇒ V = \(\frac{4}{3}\)πr³ + l³ 

From (1)

⇒ V = \(\frac{4}{3}\)πr³ + (\((\frac{S - 4\pi r^2}{6})^{\frac{3}{2}}\)

We assume V as a function of r. 

For maxima and minima,

Differentiating V again,

We get the sum of values least for r = \(\frac{1}{2}\). 

We know that diameter(d) is twice of radius. 

So, 

⇒ d = 2 x \(\frac{1}{2}\)

⇒ d = l 

∴ Thus proved.