2015 H2 Maths Paper 1, Question 4

Approach: Set up the total wire length as a constraint linking $x$ and $y$, use it to write $y$ in terms of $x$ (and $d$), then express the total area as a function of $x$ alone and maximise it.

Let the total area of the rectangle and the semicircle be $A\ \text{m}^{2}$.

The rectangle has perimeter $2x + 2y$. The semicircle, including its diameter, has perimeter (the curved arc plus the straight diameter) $\pi x + 2x$. Since the wire is cut into exactly these two pieces,

Since $y > 0$, this requires $0 < x < \frac{d}{4 + \pi}$.

The total area is the rectangle’s area plus the semicircle’s area:

Substituting the expression for $y$,

Note: the $\pi$ terms cancel exactly, so $A$ depends on $x$ but not directly on $\pi$ at this stage.

Differentiating,

At a stationary point, $\frac{dA}{dx} = 0$:

Since $\frac{d}{8} < \frac{d}{4 + \pi}$ (as $8 > 4 + \pi$), this value of $x$ lies in the valid range.

so $A$ is maximum at $x = \frac{d}{8}$.

Substituting back,

To check your answer: at $x = \frac{d}{8}$, $y = \frac{d}{2} - 2\left( \frac{d}{8} \right) - \frac{\pi}{2}\left( \frac{d}{8} \right) = \frac{d(4 - \pi)}{16} > 0$ since $\pi < 4$, confirming this is a genuine (feasible) maximum. ✓