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.

AHSEC Higher Secondary (HS) Final Examination 2019Set ANNUAL6 marks
$x^2+y^2=9$
›Reveal solutionSolution

Main: with hypotenuse $h$ fixed and one leg $x$ , maximize $A(x)=\frac12x\sqrt{h^2-x^2}$ ; the maximum occurs at $x=h/\sqrt2$ , giving equal legs. OR: find the intersection points of the circle and parabola, then integrate.

Main question. Let the hypotenuse have fixed length $h$ , and let one leg be $x$ ( $0<x<h$ ); the other leg is $\sqrt{h^2-x^2}$ .

Area: $A(x) = \dfrac12\, x\sqrt{h^2-x^2}$ . It's easier to maximize $A^2$ (same maximizer since $A>0$ ):

$A^2 = \frac14x^2(h^2-x^2) = \frac14(h^2x^2-x^4).$

$\frac{d(A^2)}{dx} = \frac14(2h^2x - 4x^3) = \frac{x}{2}(h^2-2x^2).$

Setting this to $0$ (for $x>0$ ): $h^2-2x^2=0 \implies x^2=\dfrac{h^2}{2} \implies x=\dfrac{h}{\sqrt2}$ .

Check it's a maximum: $\dfrac{d^2(A^2)}{dx^2}\Big|_{x=h/\sqrt2} = \dfrac{h^2}2-6x^2\big|_{x=h/\sqrt2} = \dfrac{h^2}2-3h^2 <0$ , confirming a maximum (also $A\to0$ at both endpoints $x=0,h$ , consistent with an interior max).

At $x=h/\sqrt2$ , the other leg is $\sqrt{h^2-h^2/2}=\sqrt{h^2/2}=h/\sqrt2$ — equal to the first leg. So the area is maximum exactly when the two legs are equal, i.e. the triangle is isosceles.

OR question. Find the intersection of $x^2+y^2=9$ (circle, radius 3) and $y^2=8x$ (parabola).

Substitute $y^2=8x$ into the circle: $x^2+8x=9 \implies x^2+8x-9=0 \implies (x+9)(x-1)=0$ . Since $x\ge0$ (parabola), $x=1$ , giving $y^2=8 \implies y=\pm2\sqrt2$ .

The smaller enclosed region is symmetric about the $x$ -axis, bounded by the parabola from $x=0$ to $1$ and by the circle from $x=1$ to $3$ . Its area is

$\text{Area} = 2\left[\int_0^1\sqrt{8x}\,dx + \int_1^3\sqrt{9-x^2}\,dx\right].$

First integral: $\displaystyle\int_0^1\sqrt8\sqrt x\,dx = \sqrt8\left[\frac23x^{3/2}\right]_0^1 = \frac{2\sqrt8}{3} = \frac{4\sqrt2}{3}.$

…

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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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