Find the maximum area of the rectangle that can be formed with fixed perimeter 20.

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Q. Find the maximum area of the rectangle that can be formed with fixed perimeter 20.

Telangana TsbieTelangana Board of Intermediate Education (Intermediate 1st Year) 2019Subjective· 7mImportance★★★★★

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Concept understanding — Maximizing Product Given Sum

The Core Intuition

You have a fixed length of rope and want the largest rectangular garden. The perimeter is fixed, so the sum of length and width is constant — but you're asked about the product of two numbers whose sum is fixed. This is the classic "Maximizing Product Given Sum" problem, appearing in optimisation, inequality proofs, and why a square beats a rectangle for area.

Suppose two numbers add up to 10:

1 and 9 → product = 9
2 and 8 → product = 16
3 and 7 → product = 21
4 and 6 → product = 24
5 and 5 → product = 25

As the numbers get closer together, the product grows; the maximum is at equality. For a fixed sum, the product is maximised when the numbers are as balanced as possible.

Note

This holds for any count of positive numbers. Three numbers summing to 30 give the maximum product when each is 10.

The Precise Statement

x_1, \dots, x_n

This follows from the AM–GM inequality:

\frac{x_1 + \cdots + x_n}{n} \geq \sqrt[n]{x_1 \cdots x_n}

with equality iff all $x_i$ are equal. Since the left side is fixed at $S/n$ , the product is bounded above by $(S/n)^n$ , achieved exactly when all numbers are equal.

Important

"Positive numbers" is crucial. If negatives are allowed, the product can be made arbitrarily large in magnitude (e.g. $x = 1000$ , $y = -990$ : sum 10, product $-990000$ ). For non-negative numbers the result holds, but the maximum is zero if any number is zero.

Why This Matters for Exams

Three main forms:

Direct: "Find two positive numbers whose sum is 20 with maximum product." → 10 and 10.
Word problems: "100 m of fencing for a rectangular pen — maximise area." Length + width = 50, so a 25 m square is best.
Inequality proofs: "For positive $a,b$ with $a+b=1$ , prove $ab \leq 1/4$ ." The two-number case.

Watch out

A common mistake: applying this to perimeter problems without halving. If all four sides sum to a fixed value, length + width is half of it. Always check what is being summed.

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Express area as a function of one side using the perimeter constraint, then maximize using the first and second derivative tests.

Let the rectangle have sides $x$ and $y$ , with perimeter $2(x+y)=20$ , so:

$x+y=10 \implies y=10-x$

Area:

$A(x) = xy = x(10-x) = 10x-x^{2}$

First derivative:

$A'(x) = 10-2x$

Setting $A'(x)=0$ : $10-2x=0 \implies x=5$ .

Second derivative test:

$A''(x) = -2 < 0$

Since $A''<0$ , $x=5$ gives 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.

CBSE 2025Set X13 marks
$x$
›Reveal solutionSolution

Maximise $xy^3$ under $x+y=60$ using the first/second derivative test; the numbers are $x=15,\ y=45$ .

Step 1 — Set up one variable.

Let the product to be maximised be $P=xy^3$ . Since $x+y=60$ , write $x=60-y$ . Then
$P(y)=(60-y)y^3=60y^3-y^4,\qquad 0<y<60.$
Step 2 — Differentiate.
$\frac{dP}{dy}=180y^2-4y^3=4y^2(45-y).$
Step 3 — Critical points.

Set $\dfrac{dP}{dy}=0$ : $4y^2(45-y)=0\Rightarrow y=0$ or $y=45$ . Since both numbers must be positive, reject $y=0$ , so $y=45$ .

Step 4 — Confirm it is a maximum.

$\frac{d^2P}{dy^2}=360y-12y^2.$ …
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CBSE 2024Set A13 marks
$x$
›Reveal solutionSolution

Express the product in one variable using $x+y=60$ , then maximise with the first derivative; the numbers are $x=15,\ y=45$ .

Concept. Reduce to a single variable via the constraint, then use $\dfrac{dP}{dy}=0$ and the second-derivative test to locate the maximum.

Let the numbers be $x$ and $y$ with $x+y=60$ , so $x=60-y$ . We maximise
$P=xy^3=(60-y)\,y^3=60y^3-y^4.$
Differentiate with respect to $y$ :

$\frac{dP}{dy}=180y^2-4y^3=4y^2(45-y).$ …
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CBSE 2022Set HE2193 marks
$x^{2/3} + y^{2/3} = 2$
›Reveal solutionSolution

Maximise $P=x(24-x)$ using calculus for Part 1; differentiate the curve implicitly and use point-slope form for Part 2 (OR).

Part 1: Let the two numbers be $x$ and $24-x$ . Their product:

$P(x) = x(24-x) = 24x-x^2$

$P'(x) = 24-2x = 0 \Rightarrow x=12$

$P''(x) = -2 < 0 \Rightarrow \text{maximum at } x=12$

The two numbers are $12$ and $24-12=12$ ; maximum product $=12\times12=144$ .

OR — Part 2: Curve $x^{2/3}+y^{2/3}=2$ at $(1,1)$ . Differentiating implicitly:

$\dfrac23 x^{-1/3} + \dfrac23 y^{-1/3}\dfrac{dy}{dx}=0 \Rightarrow \dfrac{dy}{dx} = -\left(\dfrac{y}{x}\right)^{1/3}$ …

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CBSE 2018Set ANNUAL3 marks
$Q. Find two numbers whose sum is 24 and whose product is as large as possible.$
›Reveal solutionSolution

Writing the product as $P=x(24-x)$ and maximising gives $x=12$ , so both numbers are $12$ .

Concept. To maximise a quantity subject to a constraint, express it as a one-variable function, then use $\dfrac{dP}{dx}=0$ and the second-derivative test.

Step-by-step. Let the numbers be $x$ and $24-x$ . Their product is

$P(x)=x(24-x)=24x-x^2.$

$P'(x)=24-2x.$

Set $P'(x)=0$ : $24-2x=0\Rightarrow x=12$ . …

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Q.Find the maximum area of the rectangle that can be formed with fixed perimeter 20.

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