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.

CBSE 2026Set ANNUAL1 mark
$Q. (Continuing the aluminium-box case study of Q.38) What will be the dimensions of the largest box?$
›Reveal solutionSolution
Substitute the optimal square side $x=\dfrac{2}{3}$ m (found by maximising the volume function) into the length, breadth and height expressions.

From the case study, cutting a square of side $x$ from each corner of the $3\text{ m}\times 8\text{ m}$ sheet and folding up the sides gives a box of:

Length $= (8-2x)$ m

Breadth $= (3-2x)$ m

Height $= x$ m

Maximising $V(x)=x(3-2x)(8-2x)=4x^3-22x^2+24x$ using $V'(x)=12x^2-44x+24=0$ (i.e. $3x^2-11x+6=0$ ) gives roots $x=3$ or $x=\dfrac{2}{3}$ . Since $0<x<1.5$ is required for the box to be valid, the admissible root is $x=\dfrac{2}{3}$ , and $V''\left(\tfrac{2}{3}\right)=-28<0$ confirms this is the maximum.

Substituting $x=\dfrac{2}{3}$ :

$\text{Length} = 8 - 2\left(\tfrac{2}{3}\right) = 8 - \tfrac{4}{3} = \tfrac{20}{3}\ \text{m}$ …
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CBSE 2026Set ANNUAL1 mark
$Q. (Continuing the aluminium-box case study of Q.38) What will be the side of the square removed to form the largest box?$
›Reveal solutionSolution

The optimal square side is the critical point of the volume function that lies in the valid domain and satisfies the second-derivative maximum test.

For a square of side $x$ removed from each corner of the $3\text{ m}\times 8\text{ m}$ sheet, the box volume is:

$V(x) = x(3-2x)(8-2x) = 4x^3 - 22x^2 + 24x, \qquad 0 < x < 1.5$

Differentiating and setting $V'(x)=0$ :

$V'(x) = 12x^2 - 44x + 24 = 0 \implies 3x^2 - 11x + 6 = 0$

$x = \frac{11 \pm \sqrt{121-72}}{6} = \frac{11\pm 7}{6} \implies x = 3 \text{ or } x = \frac{2}{3}$

Since the breadth $(3-2x)$ must stay positive, only $x<1.5$ is valid, so $x=3$ is rejected and $x=\dfrac{2}{3}$ is the only admissible critical point.

Confirming it is a maximum: …

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CBSE 2025Set ANNUAL1 mark
$x$
›Reveal solutionSolution

The rectangle's top corners lie on the semicircle of radius 30, so the Pythagorean relation between half the length and the breadth gives the breadth as a function of $x$ .

Place the centre $O$ of the semicircle at the origin, with the diameter along the $x$ -axis. Since the rectangle $PQRS$ is symmetric about $O$ with top side $PQ=x$ , the top corners $P,Q$ are at horizontal distance $x/2$ from $O$ . Let the breadth (height of the rectangle) be $b$ . …

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CBSE 2025Set ANNUAL1 mark
$x$
›Reveal solutionSolution

Area = length × breadth, using the breadth found in terms of $x$ .

The rectangle has length $PQ=x$ and breadth $b=\sqrt{900-x^2/4}$ (from the semicircle constraint). So the area is …

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CBSE 2025Set ANNUAL1 mark
$x$
›Reveal solutionSolution

Maximize $A(x)^2$ (equivalent and algebraically simpler) by setting its derivative to zero.

From $A(x)=x\sqrt{900-x^2/4}$ , consider $A^2 = x^2\left(900-\dfrac{x^2}{4}\right) = 900x^2-\dfrac{x^4}{4}$ (maximizing $A^2$ maximizes $A$ since $A\ge0$ ).

$\frac{d(A^2)}{dx} = 1800x - x^3 = x(1800-x^2)$

Setting this to zero: $x=0$ (rejected, gives zero area) or $x^2=1800 \Rightarrow x=\sqrt{1800}=30\sqrt2$ .

…

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CBSE 2025Set ANNUAL1 mark
$x$
›Reveal solutionSolution

Subtract the maximum rectangle area from the total semicircular field area.

Total area of the semicircular field: $\dfrac12\pi r^2 = \dfrac12\pi(30)^2 = 450\pi$ sq. m.

At maximum, $x=30\sqrt2$ , so breadth $b=\sqrt{900-\dfrac{(30\sqrt2)^2}{4}} = \sqrt{900-\dfrac{1800}{4}} = \sqrt{900-450}=\sqrt{450}=15\sqrt2$ .

…

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CBSE 2024Set ANNUAL1 mark
$R$
›Reveal solutionSolution

$r^2 = R^2 - x^2$ .

From the figure, $O$ is the sphere's centre, $C$ is the centre of the cone's circular base, $OC=x$ , $CA=r$ (radius of the cone's base), and $OA=R$ (a radius of the sphere, since $A$ lies on the sphere).

…

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CBSE 2024Set ANNUAL1 mark
$h$
›Reveal solutionSolution

$V = \dfrac{\pi}{3}(R+x)^2(R-x)$ .

From the figure, the cone's height is $h=R+x$ (from the base at $C$ up to the apex $D$ at the top of the sphere), and from the previous part, $r^2=R^2-x^2$ .

Volume of a cone: $V=\dfrac13\pi r^2h$ . …

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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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