Find the two positive numbers $x$ and $y$ such that $x + y = 60$ and $xy^3$ is maximum.

Find the two positive numbers x and y such that x + y = 60 and xy³ is maximum.

Question of 188

x

Karnataka PUCKarnataka II PUC Board 2025Subjective· 3mImportance★★★★★

0% · 0/188 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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.

--- …

Start your 14-day free trial to unlock the full solution →

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.

$\frac{d^2P}{dy^2}=360y-12y^2.$ …

Start your 14-day free trial to unlock the full solution →

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.

$\frac{d^2P}{dy^2}=360y-12y^2.$ …
Start your 14-day free trial to see the full solution →
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 $y$ :

$\frac{dP}{dy}=180y^2-4y^3=4y^2(45-y).$ …
Start your 14-day free trial to see the full solution →
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:

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

Start your 14-day free trial to see the full solution →
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 $P'(x)=0$ : $24-2x=0\Rightarrow x=12$ . …

Start your 14-day free trial to see the full solution →

←Previous chapterCh 5Continuity and Differentiability

Q.Find the two positive numbers xxx and yyy such that x+y=60x + y = 60x+y=60 and xy3xy^3xy3 is maximum.

🎓Unlock everything free for 14 days

Q.Find the two positive numbers $x$ and $y$ such that $x + y = 60$ and $xy^3$ is maximum.

Unlock everything free for 14 days