When the height of liquid in a container is $h$ m, the volume of liquid is $V\text{ m}^3$ where $V = 0.06 \left[ (4h + 1)^3 - 7 \right] \text{ and } h \ge 0.25$

1. Differentiate the expression for $V$ with respect to $h$ using the chain rule

$\frac{\text{d}V}{\text{d}h} = 0.06 \times 3(4h + 1)^2 \times \frac{\text{d}}{\text{d}h}(4h + 1)$
$\frac{\text{d}V}{\text{d}h} = 0.18(4h + 1)^2 \times 4$

2. Simplify the derivative expression

$\frac{\text{d}V}{\text{d}h} = 0.72(4h + 1)^2$

1. Find the value of $h$ when $V = 1.2$.

$1.2 = 0.06 \left[ (4h + 1)^3 - 7 \right]$
$\frac{1.2}{0.06} = (4h + 1)^3 - 7$
$20 = (4h + 1)^3 - 7$
$(4h + 1)^3 = 27$
$4h + 1 = 3$
$4h = 2 \implies h = 0.5$

2. Evaluate $\frac{\text{d}V}{\text{d}h}$ at $h = 0.5$.

From (i), $\frac{\text{d}V}{\text{d}h} = 0.06 \times 3(4h + 1)^2 \times 4 = 0.72(4h + 1)^2$
When $h = 0.5$, $\frac{\text{d}V}{\text{d}h} = 0.72(4(0.5) + 1)^2 = 0.72(3)^2 = 6.48$

3. Use the chain rule to find $\frac{\text{d}h}{\text{d}t}$.

Since the liquid leaks out, $\frac{\text{d}V}{\text{d}t} = -0.98$.
Using the chain rule: $\frac{\text{d}V}{\text{d}t} = \frac{\text{d}V}{\text{d}h} \times \frac{\text{d}h}{\text{d}t}$
$-0.98 = 6.48 \times \frac{\text{d}h}{\text{d}t}$
$\frac{\text{d}h}{\text{d}t} = -\frac{0.98}{6.48} = -\frac{49}{324} \approx -0.151$ (to 3 s.f.)

1. Interpret the negative sign and the value of the rate of change of height

The negative value of $\frac{\text{d}h}{\text{d}t}$ indicates that the height of the liquid is decreasing. The magnitude represents the rate at which it decreases.