Q.Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base.

Fix the total surface area $S$ and maximise the volume $V=\pi r^2h$ by eliminating $h$.

Step 1 — Use the constraint. $S=2\pi r^2+2\pi rh\Rightarrow h=\dfrac{S-2\pi r^2}{2\pi r}$.

Step 2 — Volume in one variable.

Step 3 — Differentiate. $\dfrac{dV}{dr}=\dfrac{S}{2}-3\pi r^2=0\Rightarrow S=6\pi r^2$. Also $\dfrac{d^2V}{dr^2}=-6\pi r<0$, a maximum. …

The idea

The surface area is fixed, which links $r$ and $h$. We use that relation to write the volume as a function of $r$ alone, then maximise with the derivative (standard CBSE method).

Set up

For a right circular cylinder with base radius $r$ and height $h$:

Work the steps

1. Solve the constraint for $h$:

2. Substitute into $V$:

3. Differentiate and find the critical point:

4. Confirm a maximum: …

Method: Fixed-Surface, Maximum-Volume Optimization for Solids of Revolution

This method solves "given a fixed total surface area, find the dimensions that maximize the volume" problems for shapes like cylinders — reducing a two-variable solid-geometry problem to a single-variable calculus problem via the surface-area constraint.

Steps

Step 1: Write both the fixed constraint and the objective in terms of the solid's dimensions

For a right circular cylinder with base radius $r$ and height $h$, closed at both ends:

Step 2: Use the constraint to eliminate one variable

Solve the surface-area equation for $h$ in terms of $r$ and the fixed constant $S$:

Step 3: Substitute into the volume formula to get a single-variable function

Step 4: Differentiate, solve for the critical radius, and confirm a maximum …

Common Mistakes

Mistake 1: Using the wrong surface-area formula (forgetting one or both circular ends)

Why it's wrong: a "given surface" cylinder problem like this one assumes a closed can with two circular ends, so $S = 2\pi r^2 + 2\pi rh$ — using only the lateral surface ($S=2\pi rh$) or only one end ($S=\pi r^2+2\pi rh$) changes every subsequent step and gives a wrong final ratio between $h$ and $r$. Correct approach: confirm whether the solid is open or closed at each end before writing the surface-area constraint, and use $2\pi r^2 + 2\pi rh$ for a fully closed cylinder.

Mistake 2: Differentiating $V$ while $h$ is still present, instead of eliminating it first

Why it's wrong: $V=\pi r^2 h$ has two independent-looking variables; differentiating with respect to $r$ while treating $h$ as constant ignores that $h$ itself depends on $r$ through the surface constraint, producing an incomplete (and wrong) derivative. Correct approach: always substitute the constraint to write $V$ as a function of $r$ alone before differentiating. …