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Is there a way to get the peak point of the following equation? $$ (a_1-a_2 x)\ln\left(1+\frac{b_1 x}{b_2 x+b_3}\right),$$ where $a_1,a_2,b_1,b_2,b_3$ are all positive constant values and $x$ is also positive. I know should make the derivative of this equation equal to $0$. But I cannot solve the derivative-equal-to-$0$ equation. Anyone can help me? Thanks very much.

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If we consider the function $$f(x)=(a_1-a_2 x)\log\left(1+\frac{b_1 x}{b_2 x+b_3}\right)$$ its derivative is $$f'(x)=\frac{{b_1} {b_3} ({a_1}-{a_2} x)}{({b_2} x+{b_3}) (x ({b_1}+{b_2})+{b_3})}-{a_2} \log \left(1+\frac{{b_1} x}{{b_2} x+{b_3}}\right)$$ and, just as you expected, there is no analytical solution for $f'(x)=0$; only numerical methods could solve the problem. For that, Newton method should be used, assuming that we have a reasonable strating guess $x_0$ of the solution.

If the solution $x_*$ of $f'(x)=0$ is obtained, eliminating the logarithm from the derivative, the extremum value of $f(x)$ would be given by $$\frac{{b_1} {b_3} ({a_1}-{a_2} x_*)^2}{{a_2} ({b_2} x_*+{b_3}) (x_* ({b_1}+{b_2})+{b_3})}$$

One point to be noticed is that $f(0)=0$ and that $f\big(\frac{a_1}{a_2}\big)=0$ which means that the extremum of the function could be between these two bounds.

At these two specific points,we also find $$f'(0)=\frac{{a_1} {b_1}}{{b_3}} >0$$ $$f'\big(\frac{a_1}{a_2}\big)=-{a_2} \log \left(1+\frac{{a_1}{b_1}}{{a_1} {b_2}+{a_2} {b_3}}\right) <0$$ Similarly, you could easily show that, at the two points, the second derivative is negative, which confirms the maximum. $$f''(0)=-\frac{{a_1} {b_1}^2}{{b_3}^2}-\frac{2 {a_1} {b_1} {b_2}}{{b_3}^2}-\frac{2 {a_2} {b_1}}{{b_3}}$$ $$f''\big(\frac{a_1}{a_2}\big)=-\frac{2 {a_2}^3 {b_1} {b_3}}{({a_1} {b_2}+{a_2} {b_3}) ({a_1} ({b_1}+{b_2})+{a_2} {b_3})}$$

Then, all the problem reduces to the numerical solution of $f'(x)=0$.

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