Rectangle with same lining for maximum discharge

Lining is on the wetted perimeter, so "same amount of lining" means the wetted perimeter is kept constant. For a fixed $P$ and $n$, $S$, discharge is maximum when the area $A$ is maximum (since $Q\propto A^{5/3}P^{-2/3}$).

Existing channel

$b=8$ m, $y=1.5$ m, $S=0.001$, $n=0.015$:

$$A=8\times1.5=12\ \text{m}^2,\quad P=8+2(1.5)=11\ \text{m},\quad R=\frac{12}{11}=1.0909\ \text{m}$$ $$Q_1=\frac1nAR^{2/3}\sqrt S=\frac{1}{0.015}(12)(1.0909)^{2/3}\sqrt{0.001}=26.81\ \text{m}^3/\text{s}$$

New dimensions for maximum discharge at $P=11$ m

$$A=by=(P-2y)y\;\Rightarrow\;\frac{dA}{dy}=P-4y=0\;\Rightarrow\;y=\frac P4=2.75\ \text{m}$$ $$b=P-2y=5.5\ \text{m}=2y\quad(\text{most economical rectangle})$$ $$A=5.5\times2.75=15.125\ \text{m}^2,\quad R=\frac y2=1.375\ \text{m}$$ $$Q_2=\frac{1}{0.015}(15.125)(1.375)^{2/3}\sqrt{0.001}=39.43\ \text{m}^3/\text{s}$$

Increase in discharge

$$\frac{Q_2-Q_1}{Q_1}\times100=\frac{39.43-26.81}{26.81}\times100=47.1\%$$

Answer: new dimensions $b=5.5$ m and $y=2.75$ m (i.e. $b=2y$); discharge increases from 26.81 to 39.43 m³/s, an increase of about 47%.