# Optimization of convex function - proof replication

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. So we have:

$$\frac{\partial \pi}{\partial d} > 0 \;\;\;\;\text{if}\;\;\;\;\; 0

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:

1. 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?
2. 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