Q.Find two numbers whose sum is $24$ and whose product is as large as possible.

Concept: For a fixed sum, the product of two numbers is maximized when the numbers are as close as possible (by AM–GM inequality).

Steps:

1. Let the numbers be $x$ and $24 - x$. Their product is $P(x) = x(24 - x) = 24x - x^2$.
2. This is a quadratic opening downward. Its maximum occurs at the vertex $x = -\frac{b}{2a} = -\frac{24}{2(-1)} = 12$. …

The idea is simple: if you have a fixed total to split into two parts, the product is largest when the parts are as balanced as possible. Why? Because the product $x(24-x)$ is a quadratic that opens downward — its peak lies exactly at the midpoint of the sum.

Let’s work through it.

1. Set up the problem. Let the two numbers be $x$ and $24-x$. Their product is

2. Recognise the shape.

   $P = -x^2 + 24x$ is a quadratic in $x$ with a negative coefficient on $x^2$. That means its graph is an upside-down parabola — it has a maximum at its vertex, not a minimum.

3. Find the vertex.

   For any quadratic $ax^2 + bx + c$, the vertex occurs at $x = -\frac{b}{2a}$. Here $a = -1$, $b = 24$, so

So the product is maximised when $x = 12$. The other number is $24 - 12 = 12$.

4. Compute the maximum product. $P_{\text{max}} = 12 \times 12 = 144.$ …

Method: Maximizing a Product for a Fixed Sum (Single-Variable Substitution)

This method handles the classic "two numbers with a fixed sum, maximize/minimize their product" family — including its geometric disguises like "fixed perimeter, maximize area."

Steps

Step 1: Name one unknown and express the other using the constraint

If the sum of two numbers is fixed at $S$, let one number be $x$; the constraint forces the other to be $S - x$. This turns a two-variable problem into a one-variable function.

Step 2: Write the quantity to optimize as a function of that one variable

Recognize this is a downward-opening quadratic — so it has a single maximum, not a minimum, at its vertex.

Step 3: Find the vertex using calculus or the vertex formula …

Common Mistakes

Mistake 1: Assuming very unequal numbers give a bigger product

Why it's wrong: intuition sometimes suggests that making one number very large and the other small should maximize the product — but the product of two numbers with a fixed sum is a downward parabola that peaks when the numbers are equal, and shrinks toward zero as they become unbalanced. Correct approach: trust the vertex/derivative calculation over the "bigger is better" instinct, and sanity-check with a quick numeric comparison (e.g. $1\times23=23$ versus $12\times12=144$).

Mistake 2: Forgetting to verify the critical point is a maximum, not a minimum …