Figure 1 shows the net of an open square-based pyramid cut from a square of cardboard of fixed side length d cm. The net consists of four congruent isosceles triangles, each with base x cm and perpendicular height $\tfrac{1}{2}d$ cm. The net is folded to form a pyramid which has a square base of side length x cm and vertical height h cm, as shown in Figure 2. Use differentiation to find the maximum volume of the pyramid in terms of d. You do not need to show that this value is a maximum volume. [The volume of a pyramid is given by $\tfrac{1}{3}\times$ base area $\times$ height.]

1. Use the geometric properties to express vertical height $h$ in terms of $x$ and $d$.

The slant height of the pyramid corresponds to the perpendicular height of the triangular flaps, which is given as $\frac{1}{2}d$.
Using Pythagoras' theorem on the right-angled triangle formed by the vertical height $h$, the half-base $\frac{x}{2}$, and the slant height $\frac{d}{2}$:
$h^2 + \left(\frac{x}{2}\right)^2 = \left(\frac{d}{2}\right)^2$
$h = \sqrt{\frac{d^2}{4} - \frac{x^2}{4}} = \frac{1}{2}\sqrt{d^2 - x^2}$

2. Formulate an expression for the volume $V$ in terms of $x$.

$V = \frac{1}{3} \times \text{base area} \times h$
$V = \frac{1}{3}x^2\left(\frac{1}{2}\sqrt{d^2 - x^2}\right)$
$V = \frac{1}{6}x^2(d^2 - x^2)^{\frac{1}{2}}$

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

Using the product rule and chain rule:
$\frac{\mathrm{d}V}{\mathrm{d}x} = \frac{1}{6} \left[ 2x(d^2 - x^2)^{\frac{1}{2}} + x^2 \left(\frac{1}{2}(d^2 - x^2)^{-\frac{1}{2}}(-2x)\right) \right]$
$\frac{\mathrm{d}V}{\mathrm{d}x} = \frac{1}{6} \left[ 2x\sqrt{d^2 - x^2} - \frac{x^3}{\sqrt{d^2 - x^2}} \right]$

4. Equate the derivative to zero to find the stationary value of $x$.

For a maximum volume, set $\frac{\mathrm{d}V}{\mathrm{d}x} = 0$:
$2x\sqrt{d^2 - x^2} - \frac{x^3}{\sqrt{d^2 - x^2}} = 0$
Multiply through by $\sqrt{d^2 - x^2}$:
$2x(d^2 - x^2) - x^3 = 0$
$2d^2x - 3x^3 = 0$
Since $x > 0$ for a valid physical length:
$x(2d^2 - 3x^2) = 0 \implies x^2 = \frac{2}{3}d^2$

5. Evaluate the maximum volume by substituting $x^2 = \frac{2}{3}d^2$ back into the volume formula.

$V_{\text{max}} = \frac{1}{6} \left( \frac{2}{3}d^2 \right) \sqrt{d^2 - \frac{2}{3}d^2}$
$V_{\text{max}} = \frac{1}{9}d^2 \sqrt{\frac{1}{3}d^2}$
$V_{\text{max}} = \frac{1}{9\sqrt{3}}d^3$
Rationalising the denominator:
$V_{\text{max}} = \frac{\sqrt{3}}{27}d^3$