What can we say about the combined area of these two shapes?

What can we say about the combined area of these two shapes?

Question

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 $x$ and hence show that $A\,\text{cm}^{2}$, the total area enclosed by the two shapes, is such that $A=\frac{\sqrt{3}x^2}{4}+\frac{(100-3x)^2}{16}.$ Calculate the value of $x$ for which $A$ has a stationary value. Determine whether this value of $x$ makes $A$ a maximum or a minimum.

Solution

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 $A\,\text{cm}^{2}$, 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.

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 $A \approx 126.4$.

$A$ 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.