Let $(q(\theta), p(\theta))$ be the mechanism. Let us show that the optimal mechanism will define a cutoff $\theta^\ast$ such that $q(\theta) = 1$ and $p(\theta) = \theta^\ast$ if and only if $\theta \ge \theta^\ast$.
The IC constraint will require that for all $\theta$ and $\theta' \in [0,1]$:
$$
\theta q(\theta) - p(\theta) \ge \theta q(\theta') - p(\theta'). \tag{IC}
$$
The PC (outside option normalized to $0$) tells us that:
$$
\theta q(\theta) - p(\theta) \ge 0. \tag{PC}
$$
Now assume that the mechanism specifies to sell to type $\theta$, so $q(\theta) = 1$
This requires that:
$$
\theta - p(\theta) \ge 0 \to p(\theta) \le \theta.
$$
Now, take any type $\theta' > \theta$. Then the (IC) constraint specifies that:
$$
\theta' q(\theta') - p(\theta') \ge \theta' - p(\theta)\ge \theta' - \theta > 0.
$$
If $p(\theta') \ge 0$ this requires that $q(\theta') = 1$.
Then using the (IC) constraint again, we have:
$$
\theta'- p(\theta') \ge \theta' - p(\theta),
$$
which implies that $p(\theta') \le p(\theta)$.
If $q(\theta) = 0$, then the (PC) constraint tells us that $p(\theta) = 0$.
Let $\theta^\ast$ be the lowest value of $\theta$ for which $q(\theta) = 1$. Summarizing, we have the following:
- If $\theta \ge \theta^\ast$, then $q(\theta) = 1$ and $p(\theta) \le p(\theta^\ast)$.
- If $\theta < \theta^\ast$ then $q(\theta) = 0$ and $p(\theta) = 0$.
- For all $\theta$ with $q(\theta) = 1$, $p(\theta) \le \theta$
Note that, as the firm tries to maximize profits, it will decide to put $p(\theta) = p(\theta^\ast) = \theta^\ast$ for all $\theta \ge \theta^\ast$.
This implies the following rule.
- For all $\theta \ge \theta^\ast$, set $q(\theta) = 1$ and $p(\theta) = \theta^\ast$.
- For all $\theta < \theta^\ast$, set $q(\theta) = 0$ and $p(\theta) = 0$.
The profits will then be equal to:
$$
\theta^\ast (1 - \theta^\ast) - \frac{1}{2}(1- \theta^\ast)^2.
$$
Taking first order conditions gives:
$$
1-2\theta^\ast + (1- \theta^\ast) = 2 - 3\theta^\ast = 0
$$
The second order condition is negative so the objective function is concave.
This gives a value of, $\theta^\ast = 2/3$.
