The picture and diagram below show a right cone-shaped magnet - NSC Technical Mathematics - Question 8 - 2023 - Paper 1

We start with the volume formula for a cone:

$V = \frac{1}{3} \pi r^2 h$

Substituting for r from the previous part:

$V = \frac{1}{3} \pi (9 - h^2) h$

Expanding this:

$V = \frac{1}{3} \pi (9h - h^3)$

Thus, we have:

$V(h) = 3h - \frac{1}{3} \pi h^3$

To find the maximum volume, we need to take the derivative of V(h) and set it to zero:

$V'(h) = 3 - \pi h^2$

Setting the derivative equal to zero for maximum:

$3 - \pi h^2 = 0$ $\pi h^2 = 3$ $h^2 = \frac{3}{\pi}$ $h = \sqrt{\frac{3}{\pi}}$

Calculating this approximation yields:

$h \approx 1.73 \text{ cm}$

Thus, the volume of the cone will be a maximum when h is approximately 1.73 cm.