Show that the surface area of a closed cuboid with square base and given volume is minimum, when it is a cube.

To demonstrate that the surface area of a closed cuboid with a square base and a fixed volume is minimized when it forms a cube, we can use calculus and some geometric reasoning.

Understanding the Problem

Let’s define the variables:

• V: Volume of the cuboid (constant)
• x: Length of one side of the square base
• h: Height of the cuboid

The volume of the cuboid can be expressed as:

V = x² * h

Surface Area Calculation

The surface area (S) of the cuboid is given by:

S = 2x² + 4xh

Expressing Height in Terms of Base Length

From the volume equation, we can express height (h) as:

h = V / x²

Substituting this into the surface area formula gives:

S = 2x² + 4x(V / x²)

This simplifies to:

S = 2x² + 4V/x

Finding the Minimum Surface Area

To find the minimum surface area, we need to take the derivative of S with respect to x and set it to zero:

S' = 4x - 4V/x²

Setting the derivative equal to zero:

4x - 4V/x² = 0

This leads to:

x³ = V

Identifying the Cube Condition

From the equation above, we find that:

x = (V)^(1/3)

Thus, when the base length x equals the cube root of the volume, the height h also equals x, confirming that the cuboid is a cube.

Conclusion

Therefore, the surface area of a closed cuboid with a square base and a fixed volume is minimized when the cuboid is a cube. This geometric property highlights the efficiency of cubes in enclosing volume with minimal surface area.