A miniature model in the shape of a triangular prism is shown above. A piece of wire, 36 cm in length, runs through all the edges of this triangular prism. The cross-section of the prism is an equilateral triangle of side $x$ cm and the length of the prism is $y$ cm.

1. Form an equation for the total length of the wire to express $y$ in terms of $x$.

The prism has 2 triangular ends (each with 3 edges of length $x$) and 3 rectangular faces (each with 1 edge of length $y$ connecting the ends).
Total length of wire $= 6x + 3y = 36$
$3y = 36 - 6x$
$y = 12 - 2x$

2. Find the area of the two triangular ends.

Area of one equilateral triangle $= \frac{1}{2}(x)(x)\sin 60^\circ = \frac{1}{2}x^2\left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{4}x^2$
Area of two triangular ends $= 2 \times \frac{\sqrt{3}}{4}x^2 = \frac{\sqrt{3}}{2}x^2$

3. Find the area of the three rectangular faces.

Area of one rectangular face $= xy$
Area of three rectangular faces $= 3xy$
Substitute $y = 12 - 2x$:
Area $= 3x(12 - 2x) = 36x - 6x^2$

4. Combine the areas to show the total surface area $A$.

$A = \frac{\sqrt{3}}{2}x^2 + 36x - 6x^2$
$A = \left(\frac{\sqrt{3}}{2} - 6\right)x^2 + 36x$
$A = \left(\frac{\sqrt{3} - 12}{2}\right)x^2 + 36x$ (Shown)

1. Differentiate $A$ with respect to $x$

$\frac{\text{d}A}{\text{d}x} = 2 \left( \frac{\sqrt{3}-12}{2} \right) x + 36 = (\sqrt{3}-12)x + 36$

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

For a stationary value, $\frac{\text{d}A}{\text{d}x} = 0$
$(\sqrt{3}-12)x + 36 = 0$
$(12-\sqrt{3})x = 36$

3. Solve for $x$ and simplify

$x = \frac{36}{12-\sqrt{3}}$
$x = \frac{36(12+\sqrt{3})}{(12-\sqrt{3})(12+\sqrt{3})}$
$x = \frac{36(12+\sqrt{3})}{144-3} = \frac{36(12+\sqrt{3})}{141} = \frac{12(12+\sqrt{3})}{47} \approx 3.51$ (to 3 s.f.)

1. Find the second derivative of $A$ with respect to $x$ to determine the nature of the stationary point.

From (i), $A = \frac{\sqrt{3}-12}{2}x^2 + 36x$.
$\frac{\text{d}A}{\text{d}x} = (\sqrt{3}-12)x + 36$
$\frac{\text{d}^2A}{\text{d}x^2} = \sqrt{3} - 12$
Since $\sqrt{3} - 12 < 0$, the stationary value of $A$ is a maximum.

2. Substitute the value of $x$ from (ii) into the expression for $A$ to find the stationary value.

From (ii), $x = \frac{36}{12-\sqrt{3}}$.
Substitute $x$ into $A$:
$A = \frac{\sqrt{3}-12}{2}\left(\frac{36}{12-\sqrt{3}}\right)^2 + 36\left(\frac{36}{12-\sqrt{3}}\right)$
$A = -\frac{12-\sqrt{3}}{2} \cdot \frac{1296}{(12-\sqrt{3})^2} + \frac{1296}{12-\sqrt{3}}$
$A = -\frac{648}{12-\sqrt{3}} + \frac{1296}{12-\sqrt{3}} = \frac{648}{12-\sqrt{3}}$

3. Rationalise the denominator to find the exact or numerical value of $A$.

$A = \frac{648(12+\sqrt{3})}{(12-\sqrt{3})(12+\sqrt{3})}$
$A = \frac{648(12+\sqrt{3})}{144-3} = \frac{648(12+\sqrt{3})}{141}$
$A = \frac{216(12+\sqrt{3})}{47} \approx 63.1 \text{ cm}^2$