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

Use the constraint to reduce to one variable, then maximise with calculus.

Step 1 — Substitute. From $x+y=60$, write $x=60-y$. Then

Step 2 — Differentiate and solve.

$P'(y)=180y^2-4y^3=4y^2(45-y)=0\Rightarrow y=45\ (y\neq 0).$ …

The idea

We have one equation linking $x$ and $y$ ($x+y=60$) and want to maximise $xy^3$. The board method is to use the constraint to eliminate one variable, turning it into a single-variable maximisation we solve with the derivative.

Set up

Since $x+y=60$ and both are positive, $x=60-y$ with $0<y<60$. The quantity to maximise becomes

Work the steps

1. Differentiate:

2. Critical points: $P'(y)=0\Rightarrow y=0$ or $y=45$. Since the numbers are positive, $y=0$ is rejected, leaving $y=45$. …

Method: Weighted Product Maximization via Substitution and Product Rule

This method handles "fixed sum, maximize a product with unequal powers" problems — e.g. maximizing $x^p y^q$ subject to $x + y = S$ — where the two variables don't appear symmetrically, so the optimum is not simply splitting the sum in half.

Steps

Step 1: Eliminate one variable using the constraint

From $x + y = S$, write $x = S - y$ (or vice versa), then substitute into the quantity to be maximized so it becomes a function of a single variable.

Step 2: Differentiate using the product rule and factor

Because the resulting expression is a product of powers of $y$, e.g. $(S-y)y^n$, differentiate using the product rule, then factor out the common power of $y$ — this is what makes solving $P'(y)=0$ tractable instead of expanding into a messy polynomial.

Step 3: Solve for the critical points, and reject any outside the valid domain …

Common Mistakes

Mistake 1: Differentiating $(60-y)y^3$ without the product rule

Why it's wrong: this expression is a product of two functions of $y$ ($60-y$ and $y^3$) — differentiating only one factor and treating the other as constant gives a completely wrong derivative and a wrong critical point. Correct approach: apply the product rule fully: $\frac{d}{dy}\big[(60-y)y^3\big] = -y^3 + (60-y)\cdot 3y^2$, then simplify.

Mistake 2: Assuming the sum should split evenly, like the equal-numbers case …