All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

\mathrm{Vol}, V = \frac{2}{3} \pi r^{3} + \frac{1}{3} \pi r^{2}h

Using Pythagoras’ Theorem, we have h^{2}=4^{2}-r^{2}

V = \frac{2}{3} \pi r^{3} + \frac{1}{3} \pi r^{2} \sqrt{4^{2}-r^{2}}

\frac{dV}{dr} = 2 \pi r^{2} + \frac{1}{3} \pi [2r\sqrt{4^{2}-r^{2}}+(r^{2})(\frac{-r}{\sqrt{4^{2}-r^{2}}})]

When r=r_{1}, \frac{dV}{dr} = 0,

2 \pi r^{2} + \frac{1}{3} \pi [2r\sqrt{4^{2}-r^{2}}-(\frac{r^{3}}{\sqrt{4^{2}-r^{2}}})] = 0

2 \pi r^{2}\sqrt{4^{2}-r^{2}} + \frac{1}{3} \pi [2r(4^{2}-r^{2})-r^{3}] = 0

2 r^{2}\sqrt{16-r^{2}} + \frac{1}{3} [32r-2r^{3}-r^{3}] = 0

2 r^{2}\sqrt{16-r^{2}} + \frac{1}{3} (32r-3r^{3}) = 0

2 r^{2}\sqrt{16-r^{2}} = \frac{1}{3} (3r^{3}-32r)

6r\sqrt{16-r^{2}} = (3r^{2}-32)

[6r\sqrt{16-r^{2}}]^{2} = (3r^{2}-32)^{2}

36r^{2}(16-r^{2}) = 9r^{4}-192r^{2}+1024


Using GC, r=1.207 \text{~or~} r=3.951

When r=1.207, using GC, \frac{dV}{dr} = 18.3 \neq 0

Thus, r=1.207 does not give a stationary value of V.

When r=3.951=r_1, ~ h=0.625

(iv) Note than r \textless 4,

Graph for 11(iv)
Graph for 11(iv)

Personal Comments:
It is rather disappointing to see some students unable to reach this question. Some panicked and differentiated wrongly, then forcefully copied what they were required to show. Students should also still continue with subsequent parts even if they cannot show (i). The interesting part of this question is that the last few parts require students to know how to use the GC effectively, in particularly setting the correct window for the graph in (iv) while (ii) and (iii) is entirely GC work as I’ve shown numerous times in classes.

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