$(1)$ \[\begin{align} F&=\rho Q\left(V-u\right)\\ &=\rho AV\left(V-u\right) \end{align}\] \[\begin{align} L&=Fu\\ &=\rho AVu\left(V-u\right) \end{align}\] $(2)$ $\frac{dL}{du}=0$のとき$L$は最大となるので \[\begin{align} \frac{dL}{du}&=\frac{d\left\{\rho AVu\left(V-u\right)\right\}}{du}+\rho AV\left\{V-u+u\cdot\left(-1\right)\right\}\\ &=\rho AV\left(V-2u\right) \end{align}\] \[\underline{\therefore \rho AV\left(V-2u\right)＝0}\] $\rho AV\ne0$だから$V-2u=0 \ \Longrightarrow\ u=\frac{V}{2}$ \[ L_{max}=\rho AV\cdot\frac{V}{2}\left(V-\frac{V}{2}\right)=\frac{\rho AV^3}{4}\] \[ \eta_{max}=\frac{最大動力値}{流体の持つ運動エネルギ}=\frac{\frac{\rho AV^{3}}{4}}{\frac{\rho AV^3}{2}}=\frac{1}{2}=0.5 \hspace{10mm} \] $(3)$ \[\begin{align} F&=\rho Q\left(V-u-\left(-\left(V-u\right)\cos\beta\right)\right)\\ &= \rho Q\left(V-u\right)\left(1+\cos\beta\right)\\ &=\rho AV\left(V-u\right)\left(1+\cos\beta\right) \end{align}\] \[\begin{align} L&=Fu\\ &=\rho AVu\left(V-u\right)\left(1+\cos\beta\right) \end{align}\]