Q.Find two positive numbers whose sum is $16$ and the sum of whose cubes is minimum.

Reduce to one variable, then minimise with calculus.

Step 1 — Substitute. Let the numbers be $x$ and $16-x$, with $0<x<16$. Minimise

Step 2 — Differentiate and solve.

Set $S'(x)=0\Rightarrow x=8$. …

The idea

The two numbers add to a fixed total, so we express both in one variable and minimise the sum of cubes using the derivative — the standard single-variable optimisation.

Set up

Let one number be $x$; the other is $16-x$, with $0<x<16$. Then

Work the steps

1. Differentiate (chain rule on the second term):

2. Solve $S'(x)=0$: $x^2=(16-x)^2\Rightarrow x^2-(16-x)^2=0.$ …

Method: Minimizing a Sum of Powers for a Fixed Total (Difference-of-Squares Shortcut)

This method handles "fixed sum, minimize (or maximize) the sum of like powers" problems, e.g. minimizing $x^3 + y^3$ subject to $x+y=S$ — the mirror image of the product-maximization family, using minimization instead.

Steps

Step 1: Reduce to one variable via the constraint

Let one number be $x$; the other is $S - x$. Write the quantity to be minimized as a single-variable function, e.g. $Q(x) = x^3 + (S-x)^3$.

Step 2: Differentiate, applying the chain rule to the second term

Step 3: Solve $Q'(x)=0$ using a difference-of-squares factoring, not expansion

Since $S \ne 0$, this reduces directly to $2x = S$, i.e. $x = S/2$ — far simpler than multiplying out the cubes. …

Common Mistakes

Mistake 1: Expecting very unequal numbers to give the minimum

Why it's wrong: because this is a minimization (not maximization), a student who has seen the "equal numbers maximize the product" pattern may wrongly assume the opposite — that unequal numbers should minimize the sum of cubes. In fact, for a convex function like $x^3+(S-x)^3$ on positive reals, the equal split is exactly what minimizes the sum, the same $x=S/2$ location as the product-maximization case, but for a structurally different reason (convexity, not concavity). Correct approach: let the calculus (second derivative sign) decide max vs. min — don't rely on the pattern from the product problem.

Mistake 2: Expanding $(S-x)^2$ fully instead of using the difference-of-squares shortcut …