Calculus

Application of Derivatives

Medium

core

Consider a closed cylinder of radius $r$ with a fixed surface area. The volume of the cylinder is maximum when its height is

$\frac{1}{3}r$

$\frac{1}{2}r$

✅ Correct Option: 4

For a closed cylinder with radius $r$ and height $h$ :

Surface Area: $S = 2\pi r^2 + 2\pi rh$

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

Since the surface area $S$ is fixed:

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

$2\pi rh = S - 2\pi r^2$

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

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

Substituting into the volume formula:

$V = \pi r^2 h$

$V = \pi r^2 \left(\frac{S}{2\pi r} - r\right)$

$V = \frac{Sr}{2} - \pi r^3$

For maximum volume, $\frac{dV}{dr} = 0$ :

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

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

$S = 6\pi r^2$

Substituting $S = 6\pi r^2$ into the surface area equation:

$6\pi r^2 = 2\pi r^2 + 2\pi rh$

$4\pi r^2 = 2\pi rh$

$h = \frac{4\pi r^2}{2\pi r}$

$h = 2r$

Therefore, the volume of the cylinder is maximum when its height is $2r$ .

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Consider a closed cylinder of radius rrr with a fixed surface area. The volume of the cylinder is maximum when its height is

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

Consider a closed cylinder of radius $r$ with a fixed surface area. The volume of the cylinder is maximum when its height is