A curve $C$ has parametric equations $x = \frac{\lambda}{1 + \lambda^3} \quad \text{and} \quad y = \frac{\lambda^2}{1 + \lambda^3}, \quad \text{for } \lambda > 0.$ The point $P$ is a variable point on $C$.

1. Form the expression for the area of rectangle $OQPR$ in terms of $\lambda$.

$A = x \times y = \left(\frac{\lambda}{1 + \lambda^3}\right) \left(\frac{\lambda^2}{1 + \lambda^3}\right) = \frac{\lambda^3}{(1 + \lambda^3)^2}$

2. Find the first derivative of the area with respect to $\lambda$ using the quotient rule.

$\begin{aligned} \frac{\text{d}A}{\text{d}\lambda} &= \frac{(1+\lambda^3)^2(3\lambda^2) - \lambda^3 \left[2(1+\lambda^3)(3\lambda^2)\right]}{(1+\lambda^3)^4} \\ &= \frac{3\lambda^2(1+\lambda^3)\left[(1+\lambda^3) - 2\lambda^3\right]}{(1+\lambda^3)^4} \\ &= \frac{3\lambda^2(1-\lambda^3)}{(1+\lambda^3)^3} \end{aligned}$

3. Equate the derivative to zero to find the stationary value of $\lambda$.

$\begin{aligned} \frac{\text{d}A}{\text{d}\lambda} &= 0 \\ \frac{3\lambda^2(1-\lambda^3)}{(1+\lambda^3)^3} &= 0 \\ 3\lambda^2(1-\lambda^3) &= 0 \end{aligned}$

Since it is given that $\lambda > 0$, $3\lambda^2 \neq 0$. Therefore:
$1 - \lambda^3 = 0 \implies \lambda = 1$

4. Perform the first derivative test to confirm that $\lambda = 1$ gives a maximum area.

Since $\lambda > 0$, both $3\lambda^2 > 0$ and $(1+\lambda^3)^3 > 0$. The sign of $\frac{\text{d}A}{\text{d}\lambda}$ strictly depends on $(1 - \lambda^3)$.

By the first derivative test, the area is a maximum when $\lambda = 1$.

1. Concisely find the value of $\lambda$ that maximises the area of rectangle $OQPR$ (as required from part a).

The area of the rectangle is $A = x \cdot y = \left(\frac{\lambda}{1 + \lambda^3}\right) \left(\frac{\lambda^2}{1 + \lambda^3}\right) = \frac{\lambda^3}{(1 + \lambda^3)^2}$.
$\frac{\text{d}A}{\text{d}\lambda} = \frac{(1 + \lambda^3)^2(3\lambda^2) - \lambda^3(2)(1 + \lambda^3)(3\lambda^2)}{(1 + \lambda^3)^4} = \frac{3\lambda^2(1 - \lambda^3)}{(1 + \lambda^3)^3}$
Setting $\frac{\text{d}A}{\text{d}\lambda} = 0$ for maximum area gives $1 - \lambda^3 = 0 \implies \lambda = 1$ (since $\lambda > 0$).

2. Differentiate the parametric equations for $x$ and $y$ with respect to $\lambda$.

Using the quotient rule:
$\frac{\text{d}x}{\text{d}\lambda} = \frac{(1+\lambda^3)(1) - \lambda(3\lambda^2)}{(1+\lambda^3)^2} = \frac{1 - 2\lambda^3}{(1+\lambda^3)^2}$
$\frac{\text{d}y}{\text{d}\lambda} = \frac{(1+\lambda^3)(2\lambda) - \lambda^2(3\lambda^2)}{(1+\lambda^3)^2} = \frac{2\lambda - \lambda^4}{(1+\lambda^3)^2}$

3. Find $\frac{\text{d}y}{\text{d}x}$ and evaluate it at $\lambda = 1$.

By the chain rule for parametric equations:
$\frac{\text{d}y}{\text{d}x} = \frac{\frac{\text{d}y}{\text{d}\lambda}}{\frac{\text{d}x}{\text{d}\lambda}} = \frac{\frac{2\lambda - \lambda^4}{(1+\lambda^3)^2}}{\frac{1 - 2\lambda^3}{(1+\lambda^3)^2}} = \frac{2\lambda - \lambda^4}{1 - 2\lambda^3}$
At the instant of maximum area ($\lambda = 1$):
$\frac{\text{d}y}{\text{d}x} = \frac{2(1) - (1)^4}{1 - 2(1)^3} = \frac{1}{-1} = -1$

4. Use connected rates of change to find the rate of change of the $y$-coordinate.

We are given the rate of change of the $x$-coordinate, $\frac{\text{d}x}{\text{d}t} = 1$. By the chain rule connecting rates of change:
$\frac{\text{d}y}{\text{d}t} = \frac{\text{d}y}{\text{d}x} \times \frac{\text{d}x}{\text{d}t}$
$\frac{\text{d}y}{\text{d}t} = (-1) \times 1 = -1$
The rate of change of the $y$-coordinate at this instant is $-1 \text{ unit per second}$.