The sum of the surface areas of a cuboid with sides `x ,2x` and `x/3` 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 sphere. Also find the minimum value of the sum of their volumes.

Explanation

To solve the problem, we need to prove that the sum of the volumes of a cuboid and a sphere is minimized when \( x \) is equal to three times the radius of the sphere. We will also find the minimum value of the sum of their volumes.

### Step 1: Calculate the Surface Areas

1. **Surface Area of the Cuboid**:
The sides of the cuboid are \( x \), \( 2x \), and \( \frac{x}{3} \).
\[
\text{Surface Area of Cuboid} = 2(lb + bh + hl) = 2\left(x \cdot 2x + 2x \cdot \frac{x}{3} + \frac{x}{3} \cdot x\right)
...