(i) Max V = $3750 \sqrt{3} \mathrm{~cm}^3$

(ii)(a)$0.125 \mathrm{~cm} \mathrm{~s}^{-1}$

(ii)(b)$=12.5 \mathrm{~cm}^2 \mathrm{~s}^{-1}$

(i) (a)

$\frac{\mathrm{d}^2 V}{\mathrm{~d} \theta^2}=5000(-4 \cos \theta \sin \theta-\sin \theta)=5000\left(-\frac{3 \sqrt{3}}{2}\right) \approx-12990<0$ when $\theta=\frac{\pi}{3}$ $V$ is a maximum when $\theta=\frac{\pi}{3}$
$\operatorname{Max} V=\left(5000 \frac{\sqrt{3}}{2}\right)\left(1+\frac{1}{2}\right)=\frac{15000 \sqrt{3}}{4}$
Maximum volume is $\frac{15000 \sqrt{3}}{\mathrm{4}}cm^3=3750 \sqrt{3} \mathrm{~cm}^3$.

(ii)(a)

Volume of water $=V=\left[\frac{1}{2} h\left(20+2 h \tan \frac{\pi}{4}\right)\right] 50$
$V=[h(10+h)] 50=500 h+50 h^2$
$\frac{d V}{d h}=500+100 h \text {. }$
When $h=3 \mathrm{~cm}, \frac{d V}{d h}=800$
$\frac{d h}{d t}=\frac{d h}{d D} \times \frac{d V}{d t}=\frac{100}{800} \mathrm{~cm} \mathrm{~s}^{-1}=\frac{1}{8} \mathrm{~cm}^{-t}=0.125 \mathrm{~cm} \mathrm{~s}^{-1}$

(ii)(b)

When the depth of the water is $h \mathrm{~cm}$, area of water surface $y=\left(10+2 h \tan \frac{\pi}{4}\right)(50)=500+100 h$ $\frac{d y}{d t}=100 \frac{d h}{d t}=\frac{d-00}{8} \mathrm{~cm}^2 \mathrm{~s}^{-1}=12.5 \mathrm{~cm}^2 \mathrm{~s}^{-1}$