A piece of wire, \(100\) cm in length, is divided into two parts. One part is bent to form an equilateral triangle of side \(x\) cm and the other is bent to form a square of side \(y\) cm.

Express \(y\) in terms of \(x\)

The perimeter of the two shapes will add up to \(100\). Hence we must have that \[ 3x+4y=100, \] since \(x\) is the side length of an equilateral triangle and \(y\) is the side length of a square. Rearranging this equation tells us that \[ y=\frac{100-3x}{4}. \]

…hence show that \(\quantity{A}{cm^2}\), the total area enclosed by the two shapes, is such that \[ A=\frac{\sqrt{3}x^2}{4}+\frac{(100-3x)^2}{16}. \]

Since the angles in an equilateral triangle are \(60^\circ\), we have that the area of the equilateral triangle is \[ \frac{1}{2}x^2\sin 60=\frac{\sqrt{3}}{4}x^2. \] The area of the square is \[ y^2=\frac{(100-3x)^2}{16}. \] Therefore \[ A=\frac{\sqrt{3}}{4}x^2+\frac{(100-3x)^2}{16} \] as required.

Calculate the value of \(x\) for which \(A\) has a stationary value.

Expanding the bracket and differentiating we have \[\begin{align*} A &= \frac{\sqrt{3}}{4}x^2 + \frac{10\,000}{16} - \frac{600x}{16} + \frac{9x^2}{16} \\ \frac{dA}{dx} &= \frac{\sqrt{3}}{2}x - \frac{600}{16} + \frac{9x}{8} \\ &=\frac{4\sqrt{3}+9}{8}x-\frac{300}{8}. \end{align*}\]

At the stationary value this is zero, \[ x=\frac{300}{4\sqrt{3}+9}\approx 18.8, \] which gives \(y \approx 10.9\), and \(A \approx 126.4\).

Determine whether this value of \(x\) makes \(A\) a maximum or a minimum.

\(A\) is a quadratic function of \(x\) so it has only one stationary value. The coefficient of \(x^2\) is positive so the stationary point must be a minimum.

Alternatively, we could evaluate the second derivative \[ \frac{d^2A}{dx^2}=\frac{4\sqrt{3}+9}{8}>0, \] for all \(x\), and so the stationary point must be a minimum.