The sum of the surface areas of the cuboid with sides `x ,\ 2x` and `x/3` and a sphere is given to be constant. Prove that the sum of the 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.

Explanation

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

### Step 1: Define the Surface Area and Volume

1. **Surface Area of the Cuboid**: The cuboid has dimensions \( x \), \( 2x \), and \( \frac{x}{3} \).
\[ \text{Surface Area of Cuboid} = 2(x \cdot 2x + x \cdot \frac{x}{3} + 2x \cdot \frac{x}{3}) = 2(2x^2 + \frac{x^2}{3} + \frac{2x^2}{3}) = 2(2x^2 + x^2) = 6x^2 \]
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