The diagram shows a shape made by cutting an equilateral triangle out of a rectangle of width $x$ cm. The perimeter of the shape is 20 cm.

1. Express the perimeter of the shape in terms of $x$ and the height of the rectangle, $y$, and solve for $y$.

Let the height of the rectangle be $y$ cm.
The perimeter consists of the bottom width ($x$), the two vertical sides ($2y$), and the two inner sides of the equilateral triangle ($2x$).
Perimeter $= x + 2y + 2x = 3x + 2y$
Given the perimeter is 20 cm:
$3x + 2y = 20$
$2y = 20 - 3x$
$y = 10 - \frac{3}{2}x$

2. Find the area of the rectangle and the area of the equilateral triangle in terms of $x$.

Area of the rectangle $= xy = x\left(10 - \frac{3}{2}x\right) = 10x - \frac{3}{2}x^2$
Area of the equilateral triangle $= \frac{1}{2}(x)(x)\sin 60^\circ = \frac{1}{2}x^2\left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{4}x^2$

3. Subtract the triangle's area from the rectangle's area to obtain the final expression for $A$.

Area of the shape, $A$ = Area of rectangle - Area of triangle
$A = 10x - \frac{3}{2}x^2 - \frac{\sqrt{3}}{4}x^2$
$A = 10x - \left(\frac{6}{4} + \frac{\sqrt{3}}{4}\right)x^2$
$A = 10x - \left(\frac{6+\sqrt{3}}{4}\right)x^2$ (Shown)

1. Differentiate the area function with respect to $x$

$\frac{\text{d}A}{\text{d}x} = 10 - 2\left(\frac{6+\sqrt{3}}{4}\right)x = 10 - \left(\frac{6+\sqrt{3}}{2}\right)x$

2. Set the derivative to zero to find the stationary point

For maximum area, $\frac{\text{d}A}{\text{d}x} = 0$
$10 - \left(\frac{6+\sqrt{3}}{2}\right)x = 0$
$\left(\frac{6+\sqrt{3}}{2}\right)x = 10$
$x = \frac{20}{6+\sqrt{3}} \approx 2.5866$

3. Verify that the stationary point is a maximum

$\frac{\text{d}^2A}{\text{d}x^2} = -\left(\frac{6+\sqrt{3}}{2}\right) < 0$
Since the second derivative is negative, the area is a maximum.

4. Calculate the maximum area and round both answers to 2 significant figures

Substitute $x = \frac{20}{6+\sqrt{3}}$ into $A$:
$A = 10\left(\frac{20}{6+\sqrt{3}}\right) - \left(\frac{6+\sqrt{3}}{4}\right)\left(\frac{20}{6+\sqrt{3}}\right)^2$
$A = \frac{200}{6+\sqrt{3}} - \frac{100}{6+\sqrt{3}} = \frac{100}{6+\sqrt{3}} \approx 12.933$

To 2 significant figures:
$x \approx 2.6$
Maximum area $\approx 13$