This is an example from the book "Game Theory and Mechanism Design", by Y. Narahari.
The buyer 1's problem becomes $$\max_{b_1\in[0,\alpha_2]}(\theta_1-b_1)\frac{b_1}{\alpha_2}.$$ How do we compute the solution of this problem? The textbook says that $b_1(\theta_1)=\theta_1/2$ if $\theta_1/2\leq\alpha_2$ and $b_1(\theta_1)=\alpha_2$ if $\theta_1/2>\alpha_2$. But I have no idea how to do this.
I would appreciate any help. Thank you.
The graph of the function $$(\theta_1-b_2)\frac{b_1}{\alpha_2}=\frac{\theta_1}{\alpha_2}b_1-\frac{1}{\alpha_2}b_1^2$$ is an upside-down parabola with maximum at $\theta_1/2.$ The quickest way to see this is either to remember the old vertex formula $-b/2a$ (for parabola $ax^2+bx+c=0$) or to take the derivative or set equal to zero.
So if $\theta_1/2 < \alpha_2,$ then the maximum of the parabola lies in the allowed interval $[0,\alpha_2],$ so $\theta_1/2$ is the location of the maximum on that interval. If $\theta_1/2 > \alpha_2$ then the maximum of the parabola is to the right of $\alpha_2,$ so the best we can do while staying in $[0,\alpha_2]$ is to take $b_1=\alpha_2.$