If the sum of the surface areas of cube and a sphere is constant, what is the ratio of an edge of the cube to the diameter of the sphere, when the sums of their volumes are minimum?

Given: sum of the surface areas of cube and a sphere is constant

To find: the ratio of an edge of the cube to the diameter of the sphere, when the sums of their volumes are minimum

Explanation: Let the side of the cube be ‘a’

Then surface area of the cube = 6a²….(i)

Let the radius of the sphere be ‘r’

Then the surface area of the sphere = 4πr²…(ii)

Now given the sum of the surface areas of cube and a sphere is constant, hence adding equation (i) and (ii), we get

6a²+4πr² = k (where k is the constant)

⇒ 6a² = k-4πr²

Now we know the volume of the cube is

V_c = a³

And volume of the sphere is

So the sum of the volumes of cube and sphere is

Now substituting equation (iii) in above equation, we get

Now we will find the first derivative of the volume, we get

Applying the sum rule of differentiation and taking out the constant terms, we get

Applying the power rule of differentiation, we get

Now to find the critical point we will equate the first derivative to 0, i.e.,

V’ = 0

⇒r = 0 or (k-4πr²) = 24r²

⇒r = 0 or k = 4πr²+24r²

⇒r = 0 or (4π+24)r² = k

Now we know, r≠0

Hence

Now we will find the second derivative of the volume equation, this can be done by again differentiating equation (ii), we get

Taking out the constant terms and applying the sum rule of differentiation, we get

Applying the product rule of differentiation, we get

Applying the power rule of differentiation, we get

Applying the differentiation, we get

Hence for , the sum of their volumes is minimum

Now substituting , in equation (iii), we get

Now we will find the ratio of an edge of the cube to the diameter of the sphere, when the sums of their volumes are minimum, i.e.,

a:2r

Hence the required ratio is

a:2r = 1:1