The sum of the surface areas of a rectangular parallelopiped with sides $x$, $2x$ and $\dfrac{x}{3}$ and a sphere is given to be constant. Prove that the sum of their volumes is minimum if $x$ is equal to three times the radius of the sphere. Also find the minimum value of the sum of their volumes.

The sum of the surface areas of a rectangular parallelopiped with sides x, 2x and (x)/(3) and a sphere is give

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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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We treat the sum of surface areas as a fixed constant, express the sphere’s radius in terms of $x$ , then write the sum of volumes as a function of $x$ alone. Using calculus (second derivative test) we show the minimum occurs when $x = 3r$ , and compute that minimum sum as $\frac{4\pi k}{9}$ , where $k$ is the constant surface area sum.

This is a classic optimization problem where two shapes share a fixed total surface area, and we want to minimise their combined volume. The key is to use the constraint to eliminate one variable, leaving a single-variable function to minimise.

1. Write the given data and the constraint

The rectangular parallelepiped has sides $x$ , $2x$ , and $\frac{x}{3}$ .

Its surface area is:

S_{\text{box}} = 2\left( x \cdot 2x + 2x \cdot \frac{x}{3} + \frac{x}{3} \cdot x \right) = 2\left( 2x^2 + \frac{2x^2}{3} + \frac{x^2}{3} \right) = 2\left( 2x^2 + x^2 \right) = 2(3x^2) = 6x^2

Let the sphere have radius $r$ . Its surface area is $S_{\text{sphere}} = 4\pi r^2$ .

The total surface area is constant; call it $k$ :

6x^2 + 4\pi r^2 = k \quad \text{(constant)}

6x^2 + 4\pi r^2 = k

2. Express $r$ in terms of $x$

From the constraint:

4\pi r^2 = k - 6x^2 \quad\Rightarrow\quad r^2 = \frac{k - 6x^2}{4\pi}

Since $r > 0$ , we need $k > 6x^2$ , which will hold for the relevant domain.

3. Write the sum of volumes

Volume of the box:

V_{\text{box}} = x \cdot 2x \cdot \frac{x}{3} = \frac{2x^3}{3}

Volume of the sphere:

V_{\text{sphere}} = \frac{4}{3}\pi r^3

So the total volume is:

V(x) = \frac{2x^3}{3} + \frac{4}{3}\pi r^3

But $r$ is not independent — substitute $r = \left( \frac{k - 6x^2}{4\pi} \right)^{1/2}$ :

V(x) = \frac{2x^3}{3} + \frac{4}{3}\pi \left( \frac{k - 6x^2}{4\pi} \right)^{3/2}

Simplify the second term:

\frac{4}{3}\pi \cdot \frac{(k - 6x^2)^{3/2}}{(4\pi)^{3/2}} = \frac{4}{3}\pi \cdot \frac{(k - 6x^2)^{3/2}}{4^{3/2} \pi^{3/2}} = \frac{4}{3} \cdot \frac{(k - 6x^2)^{3/2}}{8 \cdot \pi^{1/2}} = \frac{(k - 6x^2)^{3/2}}{6\sqrt{\pi}}

Thus:

V(x) = \frac{2x^3}{3} + \frac{(k - 6x^2)^{3/2}}{6\sqrt{\pi}}

4. Differentiate and set to zero

Differentiate with respect to $x$ :

V'(x) = 2x^2 + \frac{1}{6\sqrt{\pi}} \cdot \frac{3}{2}(k - 6x^2)^{1/2} \cdot (-12x)

Simplify:

V'(x) = 2x^2 + \frac{1}{6\sqrt{\pi}} \cdot \frac{3}{2} \cdot (-12x) \cdot (k - 6x^2)^{1/2} = 2x^2 - \frac{3x}{\sqrt{\pi}} (k - 6x^2)^{1/2}

Set $V'(x) = 0$ :

2x^2 = \frac{3x}{\sqrt{\pi}} (k - 6x^2)^{1/2}

Since $x > 0$ , divide by $x$ :

2x = \frac{3}{\sqrt{\pi}} (k - 6x^2)^{1/2}

Square both sides:

4x^2 = \frac{9}{\pi} (k - 6x^2)

Multiply through by $\pi$ :

4\pi x^2 = 9k - 54x^2

Bring terms together:

4\pi x^2 + 54x^2 = 9k \quad\Rightarrow\quad x^2(4\pi + 54) = 9k

So:

x^2 = \frac{9k}{4\pi + 54}

5. Show that $x = 3r$ at this critical point

From the constraint $6x^2 + 4\pi r^2 = k$ , substitute $k$ :

k = 6x^2 + 4\pi r^2

But from the critical condition we also have $k = \frac{(4\pi + 54)x^2}{9}$ . Equate:

6x^2 + 4\pi r^2 = \frac{(4\pi + 54)x^2}{9}

Multiply by 9:

54x^2 + 36\pi r^2 = (4\pi + 54)x^2

Cancel $54x^2$ from both sides:

36\pi r^2 = 4\pi x^2

Divide by $4\pi$ :

9r^2 = x^2 \quad\Rightarrow\quad x = 3r \quad (\text{since } x, r > 0)

This proves the required relation.

6. Verify it’s a minimum

Take the second derivative:

V''(x) = 4x - \frac{3}{\sqrt{\pi}} \left[ (k - 6x^2)^{1/2} + x \cdot \frac{1}{2}(k - 6x^2)^{-1/2} \cdot (-12x) \right]

At $x = 3r$ , using $k - 6x^2 = 4\pi r^2$ , we get $(k - 6x^2)^{1/2} = 2r\sqrt{\pi}$ . Substituting and simplifying (the algebra is straightforward but lengthy) yields $V''(x) > 0$ , confirming a minimum. …

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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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Concept understanding — Optimization Word Problem

Optimization Word Problems

The Plan of Attack

The Garden, Worked

The Common Families

Method: Optimizing a Combined Quantity for Two Shapes Sharing One Constraint

Steps

Common Mistakes

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