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

Exercise 6.3 · Q20

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.

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Concept understanding — Optimization Word Problem

Optimization Word Problems

Imagine planning a garden with 40 metres of fencing and wanting the largest rectangular area. A long, thin rectangle wastes space; a square feels roomier; somewhere in between lies the best shape. That is an optimisation problem — a fixed resource and a quantity to make as large (or as small) as possible.

Every optimisation word problem has the same skeleton: the best outcome — maximum area, minimum cost, largest volume, shortest time — under a constraint — limited material, a fixed budget, a given perimeter.

The Plan of Attack

The problem gives you a story, not a graph. Your job is to turn it into a single-variable function and find its peak or valley:

Name the quantity to optimise — call it $Q$ , and write it using variables.
Find the constraint — a relation between those variables (e.g. "perimeter $=40$ ").
Reduce to one variable — use the constraint to eliminate the rest.
Differentiate — solve $Q'(x)=0$ to find the critical points.
Confirm — use $Q''(x)<0$ for a maximum or $Q''(x)>0$ for a minimum.
Answer the question asked — give the actual dimensions/cost, not just $x$ .

Note

In board exams these problems almost always reduce to a quadratic or cubic. Once $Q(x)$ is written, the calculus is mechanical.

The Garden, Worked

40 m of fencing encloses a rectangle; maximise the area.

Objective: $A = l\,w$ .
Constraint: $2l+2w=40$ , so $l+w=20$ .
Reduce: $w=20-l$ , giving $A(l)=l(20-l)=20l-l^2$ .
Differentiate: $A'(l)=20-2l=0 \implies l=10$ .
Confirm: $A''(l)=-2<0$ , a maximum.

So $l=w=10$ m — a 10 m $\times$ 10 m square.

Watch out

A common slip: solving $A'(l)=0$ and stopping. Always check max vs min, and answer in the units asked.

The Common Families

| Problem type | Typical objective | Typical constraint |

|--------------|-------------------|--------------------| …

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Eliminating $h$ from the fixed surface area and maximising $V=\frac{Sr}{2}-\pi r^3$ gives $S=6\pi r^2$ , which forces $h=2r$ — the height equals the diameter.

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 $h$ :

\text{surface } S=2\pi r^2+2\pi rh\ (\text{fixed}),\qquad \text{volume } V=\pi r^2h.

Work the steps

Solve the constraint for $h$ :

2\pi rh=S-2\pi r^2\Rightarrow h=\frac{S-2\pi r^2}{2\pi r}.

Substitute into $V$ :

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

Differentiate and find the critical point:

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

Confirm a maximum: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL4 marks
$r$
›Reveal solutionSolution

Express the cone's base-radius in terms of its height using the sphere's geometry, write volume as a function of height alone, and maximize.

(Answering the general-students version.) Let the sphere have radius $r$ and centre $O$ , and let the cone have height $h$ and base radius $x$ . If the cone's apex and the centre are positioned so the base is at perpendicular distance $(h-r)$ from the centre, then by the Pythagorean relation on the base circle:

$x^2=r^2-(h-r)^2=2rh-h^2,\qquad 0<h<2r$

Volume: $V=\dfrac13\pi x^2h=\dfrac13\pi(2rh-h^2)h=\dfrac13\pi(2rh^2-h^3)$ .

…

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GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL4 marks
$\sqrt{2}$
›Reveal solutionSolution

Express curved surface area in terms of $r$ alone (using the fixed-volume constraint to eliminate $h$ ), then minimize.

Volume $V=\dfrac13\pi r^2h$ (fixed) $\Rightarrow h=\dfrac{3V}{\pi r^2}$ .

Curved surface area $S=\pi rl=\pi r\sqrt{r^2+h^2}$ . Work with $S^2=\pi^2r^2(r^2+h^2)=\pi^2r^4+\pi^2r^2h^2$ .

Substituting $h^2=\dfrac{9V^2}{\pi^2r^4}$ : $S^2=\pi^2r^4+\dfrac{9V^2}{r^2}$ . Let $f(r)=\pi^2r^4+\dfrac{9V^2}{r^2}$ (minimizing $S$ is equivalent to minimizing $f$ since $S>0$ ).

$f'(r)=4\pi^2r^3-\dfrac{18V^2}{r^3}=0 \Rightarrow 4\pi^2r^6=18V^2 \Rightarrow r^6=\dfrac{9V^2}{2\pi^2}$ .

…

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GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL4 marks
$\tan^{-1}\sqrt{2}$
›Reveal solutionSolution

Express the cone's volume as a function of the semi-vertical angle $\theta$ alone (slant height $l$ fixed), then maximise using calculus.

For slant height $l$ and semi-vertical angle $\theta$ : radius $r=l\sin\theta$ , height $h=l\cos\theta$ .

$V=\frac13\pi r^2h=\frac13\pi(l\sin\theta)^2(l\cos\theta)=\frac{\pi l^3}{3}\sin^2\theta\cos\theta$ .

Differentiate w.r.t. $\theta$ (product rule):

$\frac{dV}{d\theta}=\frac{\pi l^3}{3}\left[2\sin\theta\cos\theta\cdot\cos\theta+\sin^2\theta\cdot(-\sin\theta)\right]=\frac{\pi l^3}{3}\sin\theta\left(2\cos^2\theta-\sin^2\theta\right)$ .

…

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GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL4 marks
$\frac{2R}{\sqrt{3}}$
›Reveal solutionSolution

Express volume in terms of height using $r^2 = R^2 - \tfrac{h^2}{4}$ , maximise via $\tfrac{dV}{dh}=0$ .

For a cylinder of radius $r$ , height $h$ inscribed in a sphere of radius $R$ : $r^2 + \left(\tfrac h2\right)^2 = R^2 \Rightarrow r^2 = R^2 - \tfrac{h^2}{4}.$

$V = \pi r^2 h = \pi\left(R^2 - \tfrac{h^2}{4}\right)h = \pi R^2 h - \tfrac{\pi h^3}{4}.$

$\dfrac{dV}{dh} = \pi R^2 - \tfrac{3\pi h^2}{4} = 0 \Rightarrow h^2 = \dfrac{4R^2}{3} \Rightarrow h = \dfrac{2R}{\sqrt3}.$

$\dfrac{d^2V}{dh^2} = -\tfrac{3\pi h}{2} < 0$ , confirming a maximum.

…

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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.

Concept understanding — Optimization Word Problem

Optimization Word Problems

The Plan of Attack

The Garden, Worked

The Common Families

The idea

Set up

Work the steps

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

Steps

Common Mistakes

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.

The idea

Set up

Work the steps

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