[The volume of a pyramid of height $h$ is $\tfrac{1}{3}\times(\text{Base Area})\times h$.] The diagram shows a piece of paper in the shape of a circle centre $O$ and radius, $OP = 6\text{ cm}$. $ABCD$ is a square of side $2x\text{ cm}$ with the same centre $O$. $P, Q, R$ and $S$ are points on the circumference of the circle such that $AS = BS = BP = CP = CQ = DQ = DR = AR$. The shaded regions of the paper are to be cut off and the remaining paper can be folded to form a right pyramid with base $ABCD$.

1. Find the length of $OM$ and express $MP$ in terms of $x$.

Since $ABCD$ is a square of side $2x$ with centre $O$, the distance from the centre to the midpoint of a side is half the side length.
$OM = \frac{2x}{2} = x$
Since $OP$ is the radius of the circle ($6\text{ cm}$) and $O, M, P$ are collinear:
$MP = OP - OM = 6 - x$

2. Use Pythagoras' theorem to find the height of the pyramid.

When folded into a right pyramid, the apex $V$ is directly above $O$. The slant height of the pyramid is $VM = MP = 6 - x$.
In the right-angled triangle $VOM$:
$VO^2 + OM^2 = VM^2$
$h^2 + x^2 = (6 - x)^2$
$h^2 = 36 - 12x + x^2 - x^2$
$h^2 = 36 - 12x$
$h = \sqrt{4(9 - 3x)}$
$h = 2\sqrt{9 - 3x}\text{ cm}$ (shown)

1. Find the area of the square base

$\text{Base Area} = (2x)^2 = 4x^2$

2. Substitute the base area and height into the volume formula

$V = \frac{1}{3} \times \text{Base Area} \times h$
$V = \frac{1}{3} \times 4x^2 \times 2\sqrt{9-3x}$
$V = \frac{8}{3}x^2\sqrt{9-3x}$

1. Differentiate $V$ with respect to $x$ using the product rule and chain rule.

$V = \frac{8}{3}x^2(9-3x)^{\frac{1}{2}}$

$\frac{dV}{dx} = \frac{8}{3} \left[ 2x(9-3x)^{\frac{1}{2}} + x^2 \left( \frac{1}{2} \right)(9-3x)^{-\frac{1}{2}}(-3) \right]$

2. Simplify the derivative.

$\frac{dV}{dx} = \frac{8}{3} \left[ 2x(9-3x)^{\frac{1}{2}} - \frac{3}{2}x^2(9-3x)^{-\frac{1}{2}} \right]$
$\frac{dV}{dx} = \frac{8}{3} \left[ \frac{2x(9-3x) - 1.5x^2}{(9-3x)^{\frac{1}{2}}} \right]$
$\frac{dV}{dx} = \frac{8}{3} \left[ \frac{18x - 6x^2 - 1.5x^2}{(9-3x)^{\frac{1}{2}}} \right]$
$\frac{dV}{dx} = \frac{8}{3} \left[ \frac{18x - 7.5x^2}{(9-3x)^{\frac{1}{2}}} \right]$

3. Set the derivative to zero to find the stationary value of $x$.

For a stationary value, $\frac{dV}{dx} = 0$.
$\frac{8}{3} \left[ \frac{18x - 7.5x^2}{(9-3x)^{\frac{1}{2}}} \right] = 0$
$18x - 7.5x^2 = 0$
$x(18 - 7.5x) = 0$
Since $x > 0$ (as it represents a length), we have:
$18 - 7.5x = 0$
$7.5x = 18$
$x = \frac{18}{7.5} = 2.4$

1. Use the first derivative test to determine the nature of the stationary point at $x = 2.4$.

From part (iii), $\frac{dV}{dx} = \frac{4x(36-15x)}{3\sqrt{9-3x}}$ and the stationary point is at $x = 2.4$.

When $x = 2.3$, $\frac{dV}{dx} = \frac{4(2.3)(36-15(2.3))}{3\sqrt{9-3(2.3)}} > 0$.
When $x = 2.5$, $\frac{dV}{dx} = \frac{4(2.5)(36-15(2.5))}{3\sqrt{9-3(2.5)}} < 0$.

2. Conclude the nature of the stationary value.

Since $\frac{dV}{dx}$ changes from positive to negative as $x$ increases through $2.4$, the value of $x$ gives a maximum $V$.