Amongst all open (from the top) right circular cylindrical boxes of volume $125\pi$ cm$^3$, find the dimensions of the box which has the least surface area.

> [!TLDR] > To minimize the surface area of an open cylindrical box with a fixed volume, the radius and height must be equal. For a volume of $125$ cm$^3$, the dimensions are radius $\frac{5}{\sqrt[3]{\pi}}$ cm and height $\frac{5}{\sqrt[3]{\pi}}$ cm. This problem asks us to find the dimensions of an open cylindrical box that uses the least amount of material (i.e., has the least surface area) while holding a specific volume. This is a classic optimization problem in calculus. The core idea is to: 1. **Formulate:** Express the quantity to be minimized (surface area) as a function of the dimensions (radius and height). 2. **Constrain:** Use the given fixed volume to establish a relationship between the dimensions, allowing us to express one dimension in terms of the other. 3. **Reduce:** Substitute this relationship into the surface area function, making it a function of a single variable. 4. **Optimize:** Use differentiation to find the critical points of this single-variable function, which correspond to potential minimums or maximums. 5. **Verify:** Use the second derivative test to confirm that the critical point indeed yields a minimum surface area. Let's break down the solution step-by-step. 1. **Define Variables and Formulate Equations** Let $r$ be the radius of the base of the cylinder and $h$ be its height. The problem states that the box is "open from the top," meaning it has a circular base but no top lid. The volume $V$ of a right circular cylinder is given by: $$V = \pi r^2 h$$ We are given that the volume is $125$ cm$^3$. So, our constraint equation is: $$\pi r^2 h = 125 \quad \text{(Equation 1)}$$ The surface area $A$ of an open cylinder consists of the area of the circular base and the area of the curved side. Area of base $= \pi r^2$ Area of curved surface $= 2\pi rh$ So, the objective function (the quantity we want to minimize) is: $$A = \pi r^2 + 2\pi rh \quad \text{(Equation 2)}$$ 2. **Express

Amongst all open (from the top) right circular cylindrical boxes of volume 125π cm³, find the dimensions of th

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125\pi

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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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To minimize the surface area of an open cylindrical box with a fixed volume, the radius and height must be equal. For a volume of $125$ cm $^3$ , the dimensions are radius $\frac{5}{\sqrt[3]{\pi}}$ cm and height $\frac{5}{\sqrt[3]{\pi}}$ cm.

This problem asks us to find the dimensions of an open cylindrical box that uses the least amount of material (i.e., has the least surface area) while holding a specific volume. This is a classic optimization problem in calculus. The core idea is to:

Formulate: Express the quantity to be minimized (surface area) as a function of the dimensions (radius and height).
Constrain: Use the given fixed volume to establish a relationship between the dimensions, allowing us to express one dimension in terms of the other.
Reduce: Substitute this relationship into the surface area function, making it a function of a single variable.
Optimize: Use differentiation to find the critical points of this single-variable function, which correspond to potential minimums or maximums.
Verify: Use the second derivative test to confirm that the critical point indeed yields a minimum surface area.

Let's break down the solution step-by-step.

Define Variables and Formulate Equations Let $r$ be the radius of the base of the cylinder and $h$ be its height. The problem states that the box is "open from the top," meaning it has a circular base but no top lid. The volume $V$ of a right circular cylinder is given by:

V = \pi r^2 h

We are given that the volume is $125$ cm$^3$. So, our constraint equation is:

\pi r^2 h = 125 \quad \text{(Equation 1)}

The surface area $A$ of an open cylinder consists of the area of the circular base and the area of the curved side.
Area of base $= \pi r^2$
Area of curved surface $= 2\pi rh$
So, the objective function (the quantity we want to minimize) is:

A = \pi r^2 + 2\pi rh \quad \text{(Equation 2)}

Express Surface Area in Terms of a Single Variable To minimize $A$ , we need to express it as a function of a single variable, either $r$ or $h$ . We can use Equation 1 to express $h$ in terms of $r$ (or vice versa). From Equation 1:

h = \frac{125}{\pi r^2}

Now, substitute this expression for $h$ into Equation 2:

A(r) = \pi r^2 + 2\pi r \left(\frac{125}{\pi r^2}\right)

Simplify the expression for $A(r)$:

A(r) = \pi r^2 + \frac{250}{r}

This is the function we need to minimize. Note that $r$ must be positive, as it represents a physical dimension.

3. Find the Derivative of the Surface Area Function

To find the minimum surface area, we differentiate $A(r)$ with respect to $r$ and set the derivative to zero.

A'(r) = \frac{d}{dr} \left(\pi r^2 + 250r^{-1}\right)

A'(r) = 2\pi r - 250r^{-2}

A'(r) = 2\pi r - \frac{250}{r^2}

Find Critical Points Set $A'(r) = 0$ to find the critical points:

2\pi r - \frac{250}{r^2} = 0

2\pi r = \frac{250}{r^2}

Multiply both sides by $r^2$:

2\pi r^3 = 250

r^3 = \frac{250}{2\pi}

r^3 = \frac{125}{\pi}

Solve for $r$:

r = \sqrt[3]{\frac{125}{\pi}}

r = \frac{5}{\sqrt[3]{\pi}} \text{ cm}

This is the radius that potentially minimizes the surface area.

5. Verify Minimum Using the Second Derivative Test

To confirm that this critical point corresponds to a minimum, we use the second derivative test. We need to find $A''(r)$ . …

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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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Q.Amongst all open (from the top) right circular cylindrical boxes of volume 125π125\pi125π cm3^33, find the dimensions of the box which has the least surface area.

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Q.Amongst all open (from the top) right circular cylindrical boxes of volume $125\pi$ cm $^3$ , find the dimensions of the box which has the least surface area.

Unlock everything free for 14 days