Mr Alexander built a rectangular fish tank - NSC Technical Mathematics - Question 8 - 2018 - Paper 1

To find the value of x that maximizes the volume, we take the derivative of the volume function and set it to zero:

$\frac{dV}{dx} = 4.5 - 9x$

Setting the derivative equal to zero for maximization:

$4.5 - 9x = 0$

Solving for x gives: $9x = 4.5$ $x = \frac{4.5}{9} = 0.5$.

Thus, the value of x that maximizes the volume is x = 0.5.

The initial velocity of the toy car is given by evaluating the velocity function at t = 0:

$v(0) = 8 + 4(0) - (0)^2 = 8 \text{ m/s}$.

Thus, the initial velocity of the toy car is 8 m/s.

To find the velocity at t = 0.2 seconds, substitute t into the velocity function:

$v(0.2) = 8 + 4(0.2) - (0.2)^2$

Calculating this: $v(0.2) = 8 + 0.8 - 0.04 = 8.76 \text{ m/s}$.

Therefore, the velocity of the toy car at t = 0.2 seconds is 8.76 m/s.

To find the rate at which the velocity changes, we take the derivative of the velocity function:

$\frac{dv}{dt} = 4 - 2t$

Evaluating this derivative at t = 1.2 seconds:

$\frac{dv}{dt} \bigg|_{t=1.2} = 4 - 2(1.2) = 4 - 2.4 = 1.6 \text{ m/s}^2$.

Thus, the rate at which the velocity changes with respect to time when t = 1.2 seconds is 1.6 m/s².