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

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

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

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

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

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

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

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

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

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

$45r^{4}-768r^{2}+1024=0$

(ii)
Using GC, $r=1.207 \text{~or~} r=3.951$

(iii)
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 $r \textless 4$,