Fig. 1 shows the net of a triangular prism cut from a square cardboard of side length 20 cm. The net consists of one rectangle $z$ cm by $x$ cm, two rectangles $z$ cm by $y$ cm and two isosceles triangles each with base $x$ cm and perpendicular height $h$ cm. When the shaded area of the cardboard is removed, the net is folded to form a triangular prism with a rectangular base as shown in Fig. 2.

1. Relate $x$ and $y$ using the side length of the square cardboard.

Since the cardboard is a square with side length 20 cm, the total width across the three adjacent rectangles is 20 cm:
$y + x + y = 20$
$x = 20 - 2y$

2. Set up an equation using Pythagoras' theorem on the isosceles triangle.

The triangular face is isosceles with base $x$, height $h$, and slanted edge $y$ (which joins the side $y$ of the adjacent rectangle). Applying Pythagoras' theorem to half of the triangle:
$y^2 = h^2 + \left(\frac{x}{2}\right)^2$
Substituting $x = 20 - 2y$:
$y^2 = h^2 + \left(\frac{20 - 2y}{2}\right)^2$
$y^2 = h^2 + (10 - y)^2$

3. Expand the equation and solve for $y$.

$y^2 = h^2 + 100 - 20y + y^2$
$20y = h^2 + 100$
$y = \frac{100}{20} + \frac{h^2}{20}$
$y = 5 + \frac{h^2}{20} \text{ (shown)}$

4. Find $x$ in terms of $h$.

Substitute the expression for $y$ back into $x = 20 - 2y$:
$x = 20 - 2\left(5 + \frac{h^2}{20}\right)$
$x = 20 - 10 - \frac{h^2}{10}$
$x = 10 - \frac{h^2}{10}$

1. Express the dimensions $x$ and $z$ in terms of $h$ using the constraints of the 20 cm square net.

From the horizontal arrangement of the net within the 20 cm square, the total width is $y + x + y = 20 \implies x = 20 - 2y$.
Substitute $y = 5 + \frac{h^2}{20}$ (from part a):
$x = 20 - 2\left(5 + \frac{h^2}{20}\right) = 10 - \frac{h^2}{10}$
From the vertical arrangement, the total height comprises the central rectangle and the two triangles, giving $h + z + h = 20 \implies z = 20 - 2h$.

2. Formulate the volume equation and expand it to the required form.

The volume of the prism $V = (\text{Area of triangular base}) \times \text{length} = \left(\frac{1}{2}xh\right) \times z$.
Substitute $x$ and $z$:
$V = \frac{1}{2}\left(10 - \frac{h^2}{10}\right) \times h \times (20 - 2h)$
$V = \left(5 - \frac{h^2}{20}\right)(h)(20 - 2h)$
$V = \left(5h - \frac{h^3}{20}\right)(20 - 2h)$
$V = 100h - 10h^2 - \frac{20h^3}{20} + \frac{2h^4}{20}$
$V = 100h - 10h^2 - h^3 + \frac{h^4}{10}$
$V = \frac{1}{10}\left(h^4 - 10h^3 - 100h^2 + 1000h\right) \quad \text{(Shown)}$

3. Differentiate $V$ with respect to $h$.

$\frac{dV}{dh} = \frac{1}{10}\left(4h^3 - 30h^2 - 200h + 1000\right)$

4. Find the stationary values of $h$ by setting the derivative to zero.

For a maximum volume, set $\frac{dV}{dh} = 0$:
$4h^3 - 30h^2 - 200h + 1000 = 0$
Using a graphing calculator (or by factorising as $2(h - 10)(2h^2 + 5h - 50) = 0$):
$h = 10$, $h = \frac{-5 + \sqrt{425}}{4} \approx 3.90388$, or $h = \frac{-5 - \sqrt{425}}{4} \approx -6.40388$
Since length $h > 0$ and the prism length $z = 20 - 2h > 0 \implies h < 10$, we reject $h = 10$ and the negative root.
The only valid root is $h \approx 3.90388$.

5. Verify the nature of the stationary point and calculate the maximum volume.

Using the second derivative test:
$\frac{d^2V}{dh^2} = \frac{1}{10}\left(12h^2 - 60h - 200\right)$
Substitute $h \approx 3.90388$:
$\frac{d^2V}{dh^2} = \frac{1}{10}\left[12(3.90388)^2 - 60(3.90388) - 200\right] \approx -25.1 < 0$
Since $\frac{d^2V}{dh^2} < 0$, the volume is a maximum.
Substitute $h = 3.90388$ into $V$:
$V_{\text{max}} = \frac{1}{10}\left((3.90388)^4 - 10(3.90388)^3 - 100(3.90388)^2 + 1000(3.90388)\right) \approx 201.71$
The maximum volume is $202 \text{ cm}^3$ (to 3 s.f.).