Let $r$ be the radius and $h$ be the height of the right circular cylinder. Let $S$ be the given surface area (constant) and $V$ be the volume.

Surface area of cylinder:

$$S = 2\pi r^2 + 2\pi r h$$

From this, we get:

$$h = \frac{S - 2\pi r^2}{2\pi r} = \frac{S}{2\pi r} - r$$

Volume of cylinder:

$$V = \pi r^2 h$$

Substituting $h$ in terms of $r$:

$$V = \pi r^2 \left[\frac{S}{2\pi r} - r\right] = \frac{Sr}{2} - \pi r^3$$

To find maximum volume, differentiate $V$ with respect to $r$:

$$\frac{dV}{dr} = \frac{S}{2} - 3\pi r^2$$

For maxima:

$$\frac{dV}{dr} = 0$$ $$\frac{S}{2} - 3\pi r^2 = 0$$ $$S = 6\pi r^2$$

But $S = 2\pi r^2 + 2\pi r h$, therefore:

$$6\pi r^2 = 2\pi r^2 + 2\pi r h$$ $$4\pi r^2 = 2\pi r h$$ $$\mathbf{h = 2r}$$

Second derivative test:

$$\frac{d^2V}{dr^2} = -6\pi r < 0 \quad (\text{for } r > 0)$$

Since $\frac{d^2V}{dr^2} < 0$, volume is maximum when $h = 2r$.

∴ Hence, for maximum volume, height equals the diameter of the base.