The sum of the surface areas of a rectangular parallelepiped with sides x, 2x and and a sphere is given to be constant. Prove that the sum of their volumes is minimum, if x is equal to three times the radius of the sphere. Also find the minimum value of the sum of their volumes.

Given: a rectangular parallelepiped with sides x, 2x and . The sum of the surface areas of a rectangular parallelepiped and a sphere is given to be constant.

To prove: the sum of their volumes is minimum, if x is equal to three times the radius of the sphere and also find the minimum value of the sum of their volumes

Now the surface area of the rectangular parallelepiped is

Now let the radius of the sphere be r cm, then the surface area of the sphere is,

S_s=4πr²

Now sum of the surface areas of a rectangular parallelepiped and a sphere becomes,

Now given the sum of the surface areas of a rectangular parallelepiped and a sphere is constant, then

So differentiating equation (i) with respect to r, we get

Applying differentiation rule of sum, we get

Taking out the constant terms, we get

Applying the derivative, we get

Let V denotes the sum of volume of the rectangular parallelepiped and a sphere, then

For maxima or minima, the first derivative of volume should be equal to 0, i.e.,

So differentiating equation (iii) with respect to r, we get

Applying differentiation rule of sum, we get

Taking out the constant terms, we get

Applying the derivative, we get

Now substituting the value of from equation (ii), we get

i.e., x is three times the radius of the sphere when sum of the volumes of rectangular parallelepiped and sphere

Hence proved

Now to find the minimum volume of the sum of their volumes we need to find out the second derivative value at x=2r.

Hence applying derivative with respect to r to equation (iv), we get

Applying differentiation rule of sum, we get

Taking out the constant terms, we get

Applying differentiation rule of product to the first part, we get

Now substituting the value of from equation (ii), we get

Now substituting x=3r, we get

This is positive; hence V is minimum when x=3r or , and the minimum value of volume can be obtained by substituting in equation (iii), we get

Is the minimum value of the sum of their volumes.