Maximum Area with Completing the Square

You don’t always need calculus to optimise, completing the square finds the maximum or minimum of a quadratic directly. This is the intended method when the quantity to optimise is a quadratic and calculus hasn’t been assumed.

A rectangular pen is built against a straight wall, so only three sides need fencing. There are $40$ m of fencing available. Show that the area $A$ enclosed is $A = 40x - 2x^2$, where $x$ is the width, and find the maximum possible area. [6]

The working

Step 1, form the area function. Two widths ($x$ each) and one length use the fencing: $2x + \text{length} = 40$, so length $= 40 - 2x$. Then: $A = x(40 - 2x) = 40x - 2x^2 \quad \text{(M1, A1, answer shown)}$

Step 2, complete the square to find the maximum. Factor out $-2$: $A = -2(x^2 - 20x) = -2\big[(x - 10)^2 - 100\big] = -2(x - 10)^2 + 200 \quad \text{(M1, M1)}$

Step 3, read the maximum. Since $-2(x - 10)^2 \le 0$, the area is largest when the bracket is zero, at $x = 10$: $\text{maximum area} = 200 \text{ m}^2 \quad \text{(A1, A1)}$

Where the marks are won and lost

• Building the area function from the fencing constraint is the first work: length $= 40 - 2x$ (one length, two widths), not $40 - x$.
• The coefficient of $x^2$ is negative, so the vertex is a maximum, exactly what’s wanted.
• Completing the square with a factored $-2$ needs care: $-2[(x-10)^2 - 100] = -2(x-10)^2 + 200$. Forgetting to multiply the $-100$ by $-2$ is the usual slip.

Common mistakes

• Using $40 - x$ for the length (forgetting only three sides are fenced, so it’s $40 - 2x$).
• Reading the vertex as a minimum.
• Arithmetic in completing the square (the $+200$).

Full method: Maximum/Minimum & the Vertex notes. Topic home: Quadratic Functions pillar.