In $\Delta ABC$, $\angle ABC = 90^\circ$, $AB = 12\text{ cm}$ and $AC = 13\text{ cm}$. The rectangle $BPQR$ is such that its vertices $P$, $Q$ and $R$ lie on $BC$, $CA$ and $AB$ respectively. It is given that $BP = x\text{ cm}$ and $PQ = y\text{ cm}$.

1. Find the length of $BC$ using Pythagoras' theorem.

In $\Delta ABC$, $\angle ABC = 90^\circ$.
$BC = \sqrt{AC^2 - AB^2}$
$BC = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5\text{ cm}$

2. Use similar triangles to establish the relationship between $x$ and $y$.

Since $BPQR$ is a rectangle, $PQ \parallel AB$.
Therefore, $\Delta CQP$ is similar to $\Delta CAB$.
$\frac{PQ}{AB} = \frac{CP}{CB}$
Given $BP = x$, $CP = 5 - x$.
$\frac{y}{12} = \frac{5 - x}{5}$
$y = \frac{12(5 - x)}{5}$
$y = \frac{60 - 12x}{5} \text{ (shown)}$

1. Express the area of the rectangle in terms of $x$ and $y$

Area of rectangle $BPQR$, $A = BP \times PQ = x \times y$

2. Substitute the expression for $y$ from part (a) and simplify to obtain the given result

$A = x \left( \frac{60 - 12x}{5} \right)$
$A = \frac{60x - 12x^2}{5}$
$A = 12x - \frac{12x^2}{5}$ (shown)

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

$A = 12x - \frac{12}{5}x^2$

$\frac{dA}{dx} = 12 - \frac{24}{5}x$

2. Find the value of $x$ at the stationary point.

For a stationary value, $\frac{dA}{dx} = 0$.

$12 - \frac{24}{5}x = 0$

$\frac{24}{5}x = 12$

$x = 12 \times \frac{5}{24} = 2.5$

3. Calculate the stationary value of $A$.

Substitute $x = 2.5$ into $A$:

$A = 12(2.5) - \frac{12}{5}(2.5)^2$

$A = 30 - 2.4(6.25) = 30 - 15 = 15$

4. Determine the nature of the stationary value using the second derivative test.

$\frac{d^2A}{dx^2} = -\frac{24}{5} = -4.8$

Since $\frac{d^2A}{dx^2} < 0$, the stationary value is a maximum.