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 x=60−y and maximising P=60y3−y4 gives y=45, x=15, with maximum xy3=1,366,875.
The idea
We have one equation linking x and y (x+y=60) and want to maximise xy3. 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
P(y)=xy3=(60−y)y3=60y3−y4.
Work the steps
Differentiate:
P′(y)=180y2−4y3=4y2(45−y).
Critical points:P′(y)=0⇒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 xpyq 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)yn, 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 …
Mistake 1: Differentiating (60−y)y3 without the product rule
Why it's wrong: this expression is a product of two functions of y (60−y and y3) — 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: dyd[(60−y)y3]=−y3+(60−y)⋅3y2, then simplify.
Mistake 2: Assuming the sum should split evenly, like the equal-numbers case …