A craft teacher has circular sheets of card, each of radius 15 cm…

One cone. With slant height $15$ and base radius $r$, the height is $\sqrt{225 - r^2}$ and $V(r) = \dfrac{\pi}{3} r^2 \sqrt{225 - r^2}$ for $0 < r < 15$. Then $V'(r) = \dfrac{\pi}{3} \cdot \dfrac{r(450 - 3r^2)}{\sqrt{225 - r^2}}$, which is positive for $r^2 < 150$ and negative after. So $V$ is greatest at $r^2 = 150$.

(a) $r = 5 \sqrt{6} \approx 12.25\ \text{cm}$, with height $5 \sqrt{3} \approx 8.66\ \text{cm}$. The sector used has angle $360^\circ \times \dfrac{12.247}{15} \approx 293.9^\circ$, so about $294^\circ$ of the sheet makes the best cone.

(b) $V = \dfrac{\pi}{3} \times 150 \times 5\sqrt3 = 250 \sqrt{3} \pi \approx 1360.35\ \text{cm}^3$.

(c) Each semicircle gives $r = 7.5\ \text{cm}$ and height $\sqrt{225 - 56.25} \approx 12.99\ \text{cm}$. The total is $T(7.5) = 2V(7.5) \approx 1530.39\ \text{cm}^3$. Two cones from the whole sheet already beat the single best cone of (b).

(d) $T(r) = V(r) + V(15 - r)$, so $T'(r) = V'(r) - V'(15 - r)$ and $T''(r) = V''(r) + V''(15 - r)$. At $r = 7.5$ the two terms are equal, $T'(7.5) = 0$, and $T''(7.5) = 2V''(7.5) \approx 6.05\ \text{cm}$. It is positive, so the equal split is a local minimum of the total, not a maximum. Moving a little card from one cone to the other makes the total larger.

(e) The leftover sector has $r_2 = 15 - 12.247 \approx 2.753\ \text{cm}$, and its cone holds about $116.99\ \text{cm}^3$. The total is $1477.34\ \text{cm}^3$, less than in (c).

What it means. The equal split is a dip, not a peak, and the split of (e) is worse still. So the best split lies somewhere between them. A computer search over all splits puts it at $r_1 \approx 4.86\ \text{cm}$ (a sector of about $117^\circ$, the other cone taking the rest), with a total of about $1541\ \text{cm}^3$. Finding it by hand needs a polynomial equation of high degree, which is why the problem stops at (e).