Trying to solve: $$ \max_d \pi = \frac{(V^+ +2d_u + V^+\, d_u)^2}{16(V^+ - V^-)} \;\;\;\text{s.t.} \frac{4P_u}{1-d_u}>b-1 $$
And where $P_u = \frac{1}{8} (d_u-1) (V^+ d_u-V^++2 d_u)$
This is a convex function of $d$ in the restricted $(0,1)$ interval, with first derivative $\frac{(V^+-2) (V^+ (d_u-1)-2 d_u)}{8 (V^+-V^-)}$
The denominator is always positive, while the numerator is positive if $0<V^+<2$. So we have:
$$\frac{\partial \pi}{\partial d} > 0 \;\;\;\;\text{if}\;\;\;\;\; 0<V^+<2$$
and $$\frac{\partial \pi}{\partial d} < 0 \;\;\;\;\text{if}\;\;\;\;\; V^+>2$$
So this would explain the corner solution $d=0$ iff $V^+\geq 2$ but:
- The solution says $V^- < \frac{2+V^+}{2}$ and $V^+\geq 2$ are the two conditions to find a max and no solution exists outside this region. Where does $V^- < \frac{2+V^+}{2}$ come from?
- how would you go to prove that no solution exists outside this range $V^+>2$?
question 2: potential proof In particular $d=1$ seems to be a candidate when $V^+$ is not so large, given a positive first derivative $d \to 1$ and the constraint would be violated: $$\frac{4\left(\frac{1}{8} (1-d_u) (-V^+ d_u+V^+-2 d_u)\right)}{(1-d_u)^2} = \to \frac{\frac12 (-V^+ d_u+V^+-2 d_u)}{(1-d_u)} = \frac{-2 }{0} \to \infty $$ So $d=1$ is not an eq. Is there a more general way to show this?
BACKGROUND INFO:
I am a bit baffled by a statement in this article. At pag. 7 of the pdf they write:
"plugging optimal price pu in (5) into profits in (4) obtains monopoly profits as a function of disclosure. Maximizing [it] with respect to dm subject to the constraint $d_m \geq 0$ (feasible disclosure range) and $v_f = \frac{4p_u}{(1-d)^2} > V^-$ (uncovered market) obtains optimal disclosure in an uncovered market $d_u=0$"
Now I had tried to replicate this result that should be quite immediate. By simply taking (5) and putting it into (4) I get the $\pi$ experession
