Q.Find two positive numbers $x$ and $y$ such that their sum is 35 and the product $x^2y^5$ is a maximum.

Reduce to one variable using the constraint, then maximise.

Step 1 — Substitute. From $x+y=35$, put $y=35-x$. Then

Step 2 — Differentiate (product rule):

$P'(x)=2x(35-x)^5+x^2\cdot 5(35-x)^4(-1)=x(35-x)^4\big[2(35-x)-5x\big]=x(35-x)^4(70-7x).$ …

The idea

One constraint links $x$ and $y$, so we eliminate a variable and maximise a single-variable function with the derivative — the standard CBSE approach.

Set up

From $x+y=35$ with $x,y>0$, take $y=35-x$, $0<x<35$. Then

Work the steps

1. Differentiate with the product rule:

Factor out $x(35-x)^4$:

2. Critical points: $P'(x)=0$ gives $x=0$, $x=35$, or $70-7x=0\Rightarrow x=10$. The endpoints $x=0,35$ make the product zero (not a maximum), so the interior critical point is $x=10$. …

Method: Weighted Product Maximization — Factoring a Mixed-Power Derivative

This method extends the fixed-sum product-maximization technique to expressions where both variables carry unequal exponents, e.g. maximizing $x^m y^n$ subject to $x + y = S$.

Steps

Step 1: Substitute the constraint to get a single-variable function

Replace one variable using $y = S - x$ (or the reverse), giving $P(x) = x^m (S-x)^n$.

Step 2: Differentiate with the product rule, keeping both terms symbolic before simplifying

Step 3: Factor out the common powers $x^{m-1}(S-x)^{n-1}$

This is the key algebraic move: after factoring, what's left inside the brackets is a simple linear expression in $x$, which is much easier to solve than the unfactored derivative. …

Common Mistakes

Mistake 1: Arithmetic slip expanding the bracket $2(35-x) - 5x$

Why it's wrong: this is where the whole problem's answer lives — a sign or distribution error here (e.g. writing $70 - 2x - 5x$ as $70+2x-5x$) changes the critical point entirely. Correct approach: expand carefully term by term: $2(35-x) - 5x = 70 - 2x - 5x = 70 - 7x$.

Mistake 2: Misapplying the product rule across the two factors

Why it's wrong: $x^2(35-x)^5$ has two factors, but a student differentiating quickly might forget the chain rule's extra $(-1)$ from differentiating $(35-x)^5$, or drop one of the two product-rule terms altogether — either error changes the factored form and the resulting critical point. Correct approach: write out both product-rule terms explicitly before factoring: $2x(35-x)^5 + x^2\cdot 5(35-x)^4\cdot(-1)$.

Mistake 3: Choosing the wrong critical point out of $x=0$, $35$, $10$ …