The height of a closed cylinder of given surface area and maximum volume is equal to the
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Double the radius of the base
Let the radius be r and the height h. The total surface area S of a closed cylinder is fixed:
\( \displaystyle S=2\pi r^2+2\pi rh \)
\( \displaystyle h=\frac{S-2\pi r^2}{2\pi r} \)
In terms of r only, the volume is
\( \displaystyle V=\pi r^2h \)
\( \displaystyle =\pi r^2\cdot\frac{S-2\pi r^2}{2\pi r} \)
\( \displaystyle =\frac{Sr}{2}-\pi r^3 \)
\( \displaystyle \frac{dV}{dr}=\frac{S}{2}-3\pi r^2 \)
\( \displaystyle \frac{dV}{dr}=0\ \Rightarrow\ S=6\pi r^2 \)
\( \displaystyle \frac{d^2V}{dr^2}=-6\pi r\lt 0,\ \text{so V is maximum} \)
Putting \( S=6\pi r^2 \) back into the surface area:
\( \displaystyle 2\pi r^2+2\pi rh=6\pi r^2 \)
\( \displaystyle 2\pi rh=4\pi r^2 \)
\( \displaystyle h=2r \)
The height equals the diameter, that is, double the radius of the base.
Why the other options are incorrect
Option (1) Radius of the base — Incorrect. h = r is the result for a cylinder open at the top (surface area \( \pi r^2+2\pi rh \)), not for a closed one.
Option (2) Half of radius of the base — Incorrect. At maximum volume \( S=6\pi r^2 \), which forces h = 2r, not \( \frac{r}{2} \).
Option (3) Two third of radius of the base — Incorrect. At maximum volume \( S=6\pi r^2 \), which forces h = 2r, not \( \frac{2r}{3} \).
NCERT (2026-27 edition):
Maxima and Minima
