The volume of liquid in a container, $V\ \text{m}^3$, is given by $V = 0.05\bigl[(3x+2)^3 - 8\bigr],$ where $x$ is the height of the liquid in metres. The liquid enters the container at a constant rate of $0.081\ \text{m}^3/\text{s}$.

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

$\dfrac{dV}{dx} = 0.05 \times 3(3x+2)^2 \times \dfrac{d}{dx}(3x+2)$

2. Simplify the derivative

$\dfrac{dV}{dx} = 0.05 \times 3(3x+2)^2 \times 3 \\ = 0.45(3x+2)^2$

1. Substitute $V = 0.95$ into the given equation for $V$

$0.95 = 0.05\bigl[(3x+2)^3 - 8\bigr]$
$\frac{0.95}{0.05} = (3x+2)^3 - 8$
$19 = (3x+2)^3 - 8$

2. Solve for $x$

$(3x+2)^3 = 27$
$3x+2 = \sqrt[3]{27}$
$3x+2 = 3$
$3x = 1$
$x = \frac{1}{3}$

1. Find the value of $x$ and $\frac{dV}{dx}$ when $V = 0.95$

From part (b), when $V = 0.95$, $x = \frac{1}{3}$.
From part (a), $\dfrac{dV}{dx} = 0.45(3x+2)^2$.
Substitute $x = \frac{1}{3}$ into $\dfrac{dV}{dx}$:
$\dfrac{dV}{dx} = 0.45\left(3\left(\frac{1}{3}\right)+2\right)^2 = 0.45(3)^2 = 4.05$

2. Apply the chain rule to find $\frac{dx}{dt}$

By the chain rule, $\dfrac{dV}{dt} = \dfrac{dV}{dx} \times \dfrac{dx}{dt}$.
We are given that the liquid enters at a constant rate, so $\dfrac{dV}{dt} = 0.081$.
$0.081 = 4.05 \times \dfrac{dx}{dt}$
$\dfrac{dx}{dt} = \dfrac{0.081}{4.05} = 0.02$

3. Explain the significance of the answer

The value $0.02$ means that the height of the liquid in the container is increasing at a rate of $0.02\text{ m/s}$ at the instant when the volume is $0.95\text{ m}^3$.