Liquid is poured, at a constant rate of 15 cm$^3$/s, into a container. When the depth of liquid in the container is $x$ cm, the volume, $V$ cm$^3$, of the liquid in the container is given by
$V = \frac{1}{4}x(x + 16).$ Find, when $V = 20$,

1. Substitute $V = 20$ into the given equation and rearrange into standard quadratic form

$20 = \frac{1}{4}x(x + 16)$
$80 = x^2 + 16x$
$x^2 + 16x - 80 = 0$

2. Solve the quadratic equation and reject the invalid root

$(x + 20)(x - 4) = 0$
$x = -20 \text{ or } x = 4$
Since depth $x$ must be positive, $x = 4$.

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

$V = \frac{1}{4}x^2 + 4x$

$\frac{dV}{dx} = \frac{1}{2}x + 4$

2. Evaluate $\frac{dV}{dx}$ at $x = 4$ (from part i)

When $x = 4$,
$\frac{dV}{dx} = \frac{1}{2}(4) + 4 = 6$

3. Apply the chain rule to find the rate of change of depth

Using the chain rule, $\frac{dV}{dt} = \frac{dV}{dx} \times \frac{dx}{dt}$

$15 = 6 \times \frac{dx}{dt}$

$\frac{dx}{dt} = \frac{15}{6} = 2.5$