We need to deal with the mechanism of competetive inhibition+substrate inhibition:
\begin{align}
  \ce{E + S  <=> ES}   \quad &K_\text{M} = \frac{\ce{[E]}\ce{[S]}}{\ce{[ES]}} \tag{R1/1} \\
  \ce{ES     -> E + P} \quad &r = k[\ce{ES}] \tag{R2/2} \\
  \ce{E + I  <=> EI}   \quad &K_\text{I} = \frac{\ce{[E]}\ce{[I]}}{\ce{[EI]}} \tag{R3/3} \\
  \ce{ES + S <=> ES2}  \quad &K_\text{S} = \frac{\ce{[ES]}\ce{[S]}}{\ce{[ES2]}} \tag{R4/4} 
\end{align}
where:
- $\ce{E}$ is the enzyme.
- $\ce{ES}$ is the enzyme-substrate complex.
- $\ce{I}$ is the inhibitor.
- $\ce{EI}$ is the enzyme-inhibitor complex.
- $\ce{ES2}$ is the compound generated by the reaction of the enzyme and the enzyme-substrate complex, that is taking away the substrate the enzyme needs.
- The $K$'s are the equilibrium constants, e.g., $K_\text{M}$ is the Michaelis's one. In biochemistry, it is standard practice to write them upside down.

Our task is to find an expression for the rate of formation of the product $\ce{P}$, in terms of the substrate concentration $\ce{S}$, the variable we can easily track.

**Enzyme balance** The initial concentration of enzyme is equal to
\begin{align}
  C_\text{E0} &= \underbrace{[\ce{E}]}_1 +
                 [\ce{ES}] +
                 \underbrace{[\ce{EI}]}_3 +
                 \underbrace{[\ce{ES2}]}_4 \tag{5} \\
\end{align}
now we combine underbrace $1$ with Eq. (1), underbrace $3$ with Eq. (3), and underbrace $4$ with Eq. (4)
\begin{align}
  C_\text{E0} &= [\ce{E}] + [\ce{ES}] + [\ce{EI}] + [\ce{ES2}] \\
  C_\text{E0} &= \frac{[\ce{ES}]}{[\ce{S}]}K_\text{M} + [\ce{ES}] +
                 \frac{\overbrace{[\ce{E}]}^1}{[\ce{I}]}K_\text{I} +
                 \frac{[\ce{ES}][\ce{S}]}{K_\text{S}} \tag{6} \\
\end{align}
Again, we combine overbrace $1$ with Eq. (1), because $[\ce{E}]$ appears once more
\begin{align}
  C_\text{E0} &=
  \color{blue}{\frac{[\ce{ES}]}{[\ce{S}]}K_\text{M}} + [\ce{ES}] +
  \color{blue}{\frac{[\ce{ES}]}{[\ce{S}]}K_\text{M} \left(\frac{K_\text{I}}{[\ce{I}]}\right)} +
  \frac{[\ce{ES}][\ce{S}]}{K_\text{S}} \\
  C_\text{E0} &=
  \frac{[\ce{ES}]}{[\ce{S}]}K_\text{M}
  \underbrace{\left(1 + \frac{K_\text{I}}{[\ce{I}]}\right)}_\alpha
  + [\ce{ES}] +
  \frac{[\ce{ES}][\ce{S}]}{K_\text{S}} \tag{7} \\
\end{align}
The main issue with Eq. (7), is that $\ce{[I]}$ is still there. We could use an inhibitor balance, but unfortunately, we can prove that we can't have a simple final expression. If you want further math just say in the comments. I will name underbraced part with $\alpha$, and name it the **inhibitor power**. We will explore its consequences further. Eq. (7) now becomes
\begin{align}
  C_\text{E0} &=
  \frac{[\ce{ES}]}{[\ce{S}]}K_\text{M} \alpha +
  [\ce{ES}] +
  \frac{[\ce{ES}][\ce{S}]}{K_\text{S}} \\
  C_\text{E0} &= [\ce{ES}] \left(
  \frac{K_\text{M} \alpha}{[\ce{S}]} + 1 + \frac{[\ce{S}]}{K_\text{S}}\right) \\
  C_\text{E0} &= [\ce{ES}] \left(
  \frac{K_\text{M}\alpha + [\ce{S}]}{[\ce{S}]} + \frac{[\ce{S}]}{K_\text{S}}\right) \\
  C_\text{E0} &= [\ce{ES}] \left(
  \frac{K_\text{S}K_\text{M}\alpha + K_\text{S}[\ce{S}] + [\ce{S}]^2}
  {K_\text{S}[\ce{S}]}\right)\\
  [\ce{ES}]  &=  \left(\frac{K_\text{S}[\ce{S}]}
  {K_\text{S}K_\text{M}\alpha + K_\text{S}[\ce{S}] + [\ce{S}]^2}\right)C_\text{E0}
  \tag{7} \\
\end{align}
Combining Eqs. (2) and (7)
\begin{equation}
  r = \frac{kK_\text{S}[\ce{S}]C_\text{E0}}
  {K_\text{S}K_\text{M}\alpha + K_\text{S}[\ce{S}] + [\ce{S}]^2} \rightarrow
  \boxed{\frac{r}{kC_\text{E0}} = \frac{[\ce{S}]}
  {K_\text{M}\alpha + [\ce{S}] + \dfrac{[\ce{S}]^2}{K_\text{S}}}} \tag{8} \\
\end{equation}
Eq. (8) is exactly what we want, a rate which is only a function of the substrate concentration. I also put it in a dimensionless form, which will be easier to analyze. The plot of Eq. (8) is the following, where I employed for simplicity $K_\text{S} = K_\text{M} = 1 \; \pu{mol/dm3}$:

[![enter image description here][1]][1]

Observations:
- If $\alpha = 1$, the inhibitor power is zero, and $\ce{R3}$ just vanishes. We end with only substrate inhibition.
- As $\alpha$ goes up, $\ce{R3}$ is more aggressive. This means that more enzime is consumed and be inactivated. Thus, the reaction goes down for every concentration of substrate.
- Any operation with too high concentrations of a substrate is discouraged, since $\ce{R4}$ is favored, and the substrate inhibition is too predominant.
- For all cases, we have an optimum value of substrate concentration, where the rate is maximum. This is really obvious in Eq. (8). In the numerator we have a linear function, that will win when $\ce{[S]}$ is low, and in the denominator a quadratic function that will when when $\ce{[S]}$ is high. This is a typical characteristic that substrate inhibition displays. We can prove that this happens in
$$ \frac{\mathrm{d}(r/kC_\text{E0})}{\mathrm{d}[S]} = 0 \rightarrow
\boxed{[\ce{S}]_\text{opt} = \sqrt{K_\text{S}K_\text{M}\alpha}} \tag{9} $$
This is the curve in magenta, thus, the best "ideal" operation will be the one that follows this curve for every instant in time in the biochemical reactor, disregarding the spatial effects.


  [1]: https://i.sstatic.net/e2Rsz.jpg