Solution — 2025 Question 4 Differentiation & Applications

(a)

$\begin{matrix} 2 x^{2} + 4 x y & = 96 \\ x^{2} + 2 x y & = 48 \\ 2 x y & = 48 - x^{2} \\ y & = \frac{48 - x^{2}}{2 x} \end{matrix}$

[use the area of the cuboid to find y in terms of x]

$\begin{matrix} V & = x^{2} y \\ = x^{2} (\frac{48 - x^{2}}{2 x}) \\ = \frac{x (48 - x^{2})}{2} \\ = \frac{48 x - x^{3}}{2} \\ = 24 x - \frac{1}{2} x^{3} ∎ \end{matrix}$

[find an expression for the volume]

(b)

$\begin{matrix} V & = 24 x - \frac{1}{2} x^{3} \\ \frac{d V}{d x} & = 24 - \frac{3}{2} x^{2} \end{matrix}$

[differentiate the volume to find dV/dx]

At maximum volume,

$\begin{matrix} \frac{d V}{d x} & = 0 \\ 24 - \frac{3}{2} x^{2} & = 0 \\ 48 - 3 x^{2} & = 0 \\ 3 x^{2} & = 48 \\ x^{2} & = 16 \\ x & = 4 \end{matrix}$

[solve for x when dV/dx = 0]

Differentiating again,

$\begin{matrix} \frac{d V}{d x} & = 24 - \frac{3}{2} x^{2} \\ \frac{d^{2} V}{d x^{2}} & = - 3 x \\ \frac{d^{2} V}{d x^{2}} & < 0 when x = 4 \end{matrix}$

[differentiate again to find d^2V/dx2]

Hence $V$ is a maximum when $x = 4 ∎$

$\begin{matrix} Maximum volume \\ = 24 (4) - \frac{1}{2} {(4)}^{3} \\ = 64 ∎ {cm}^{3} \end{matrix}$

[finding the maximum volume of the cuboid]