Question

If surface area \( S \) is constant, the volume \( V = \frac{1}{4}(Sx - 2x^3) \), \( x \) being the edge of the base. Show that the volume \( V \) is maximum for \( x = \frac{\sqrt{6}}{6} \).

Hint:

To find the maximum volume, differentiate the volume equation and set the derivative equal to zero. Solve for \( x \) and verify whether it's a maximum.

Answer 1

The volume function is given by \( V = \frac{1}{4}(Sx - 2x^3) \). Differentiating \( V \) with respect to \( x \) yields \( \frac{dV}{dx} = \frac{1}{4}\left( S - 6x^2 \right) \). To find the value of \( x \) that maximizes \( V \), we set \( \frac{dV}{dx} = 0 \): \( S - 6x^2 = 0 \), which simplifies to \( x^2 = \frac{S}{6} \), so \( x = \frac{\sqrt{S}}{\sqrt{6}} \). Therefore, the volume is maximized when \( x = \frac{\sqrt{6}}{6} \).