Consider a closed cylinder of radius r with a fixed surface area. The…

The correct answer is Option (4) → \(2r\)

Let S be the fixed total surface area and h the height. For a closed cylinder:

    \(S = 2\pi r^2 + 2\pi rh\)

    \( \displaystyle h = \frac{S - 2\pi r^2}{2\pi r} \)

The volume as a function of r:

    \( \displaystyle V = \pi r^2 h = \frac{r(S - 2\pi r^2)}{2} = \frac{Sr}{2} - \pi r^3 \)

    \( \displaystyle \frac{dV}{dr} = \frac{S}{2} - 3\pi r^2 \)

Putting \(\frac{dV}{dr} = 0\):

    \(S = 6\pi r^2\)

    \( \displaystyle \frac{d^2V}{dr^2} = -6\pi r \lt 0 \)

so this gives the maximum volume. Using \(S = 6\pi r^2\) in the surface area:

    \(2\pi r^2 + 2\pi rh = 6\pi r^2\)

    \(2\pi rh = 4\pi r^2\)

    \(h = 2r\)

Tip: A result worth remembering: a closed cylinder of given surface area has the largest volume when its height equals its diameter.

NCERT (2026-27 edition):
Maxima and Minima