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
Maximizing Product Given Sum
For positive reals x1,…,xn with fixed sum S, the product x1x2⋯xn is maximised when all are equal:
x1=x2=⋯=xn=nS
This follows from the AM–GM inequality:
nx1+⋯+xn≥nx1⋯xn
with equality iff all xi 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≤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.
Substituting y=35−x and maximising P=x2(35−x)5 gives x=10, y=25; the maximum of x2y5 is 102⋅255.
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
P(x)=x2y5=x2(35−x)5.
Work the steps
Differentiate with the product rule:
P′(x)=2x(35−x)5+x2⋅5(35−x)4⋅(−1).
Factor out x(35−x)4:
P′(x)=x(35−x)4[2(35−x)−5x]=x(35−x)4(70−7x).
Critical points:P′(x)=0 gives x=0, x=35, or 70−7x=0⇒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 xmyn 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)=xm(S−x)n.
Step 2: Differentiate with the product rule, keeping both terms symbolic before simplifying
P′(x)=mxm−1(S−x)n+xm⋅n(S−x)n−1⋅(−1)
Step 3: Factor out the common powers xm−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. …
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: x2(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+x2⋅5(35−x)4⋅(−1).
Mistake 3: Choosing the wrong critical point out of x=0, 35, 10 …