Profit maximization question I have the following question 

A monopolist has a cost function, $\textrm{c}(q)=q$. It faces the following demand function:
  $$ q(p) = \begin{cases} \frac{100}{p} & \textrm{if } p \leq 20 \\
0 & \textrm{if } p > 20 \end{cases} $$
  What are the profit-maximizing price and output?
A monopolist has a cost function, $c(q)=\alpha q$, where $\alpha$ is its fixed marginal cost. Its demand function has a constant price elasticity of demand whose value is $-3$. The government imposes a tax of $6 per unit of output. By how much (in dollars) will the monopolist's price rise?

My solution is 
For first part:
I found $\textrm{revenue} = \left(\frac{100}{p}\right) \cdot p = 100 $
$\textrm{Price} = 20$ 
$\textrm{Output} = 5$ 
So $\textrm{maximum profit} = 100-5 = 95$. 

For the second part
calculate price eleasticity
By the definition of price elasticity $\frac{\partial q/\partial p}{q/p}$
$$-3=(-100/p^2)*(p/q)$$
$$q=100/3p$$
For price $p<20$ 
Revenue = 100/3
Cost function = $\alpha q -6q=(\alpha -6)(100/3p)$ 
In order not to change maximum profit 
$95=(100/3)-(100/3p)(\alpha -6)$
$2.85=1-((\alpha -6)/p)$
$$p= (6-\alpha)/1.85$$
But I am not sure my solution. Please help me thank you. 
What is the value of alpha? And I am not sure in the last part, maximum profit is equal to 95 again or not. 
 A: @callculus's comments are helpful for this problem. The answer in this post would be based on these comments (yet with some necessary improvements). Also, for the second part of your question, I would like to understand it in this way: what is the price difference before and after the imposition of tax if the monopolist always chooses to take profit-maximized price.
Profit-maximized price and output
Since
$$
\text{profit}=\text{revenue}-\text{cost}=\text{demand}\times\text{price}-\text{cost},
$$
we have in general,
$$
\text{profit}=q(p)p-c(q)=q(p)p-q=q(p)\left(p-1\right).
$$
We are expected to find an optimal $p$ such that the profit is maximized.
As per the definition of $q=q(p)$, when $p\le 20$,
$$
\text{profit}=\frac{100}{p}\left(p-1\right)=100\left(1-\frac{1}{p}\right).
$$
Obviously, in this case, $p=20$ maximizes the profit, giving its value of $95=100\left(1-1/20\right)$.
When $p>20$,
$$
\text{profit}=0\left(p-1\right)=0,
$$
which is constantly smaller than the maximal profit in the last case. Therefore, this case could be dropped without any further consideration.
To sum up, $p=20$ is the optimal price, leading to the demand of $5$ and profit of $95$.
Increase in the monopolist's price
Since the demand $q$ observes a constant price elasticity of demand whose value is $-3$, we have by definition (here $\log$ means the natural logarithm, to the base $e$),
$$
\text{elasticity}=\frac{{\rm d}q/q}{{\rm d}p/p}=-3\iff\frac{{\rm d}q}{q}=-3\frac{{\rm d}p}{p}\iff{\rm d}\log q=-3{\rm d}\log p={\rm d}\log p^{-3}.
$$
Therefore, we obtain the dependence relation between $q$ and $p$:
$$
q=q(p)=\frac{\beta}{p^3},
$$
where $\beta>0$ is a fixed constant (this constant acts, mathematically, as the constant of integration).
Let $r\ge 0$ be the tax in dollar per unit output.
Now, we have
$$
\text{profit}=q(p)p-c(q)-rq=\frac{\beta}{p^3}p-\alpha q-rq=\frac{\beta}{p^3}p-\alpha\frac{\beta}{p^3}-r\frac{\beta}{p^3}=\frac{\beta}{p^2}\left(1-\frac{\alpha+r}{p}\right).
$$
To maximize the profit, we need to, mathematically, maximize
$$
\frac{\beta}{p^2}\left(1-\frac{\alpha+r}{p}\right).
$$
Yet since $\beta>0$ is a fixed constant, it suffices to maximize
$$
\frac{1}{p^2}\left(1-\frac{\alpha+r}{p}\right).
$$
Define
$$
f(p)=\frac{1}{p^2}\left(1-\frac{\alpha+r}{p}\right),
$$
and we have, after some tedious calculation and simplification,
$$
f'(p)=\frac{3\left(\alpha+r\right)-2p}{p^4}.
$$
This implies that
$$
p=\frac{3}{2}\left(\alpha+r\right)
$$
is the optimal price, because it is the only value that satisfies $f'(p)=0$.
Therefore, when the government imposes no tax, we have $r=0$, and the optimal price is
$$
\frac{3}{2}\alpha.
$$
By contrast, when the government imposes a tax of $r_0=\$6$ per unit output, we have $r=r_0$, and the optimal price becomes
$$
\frac{3}{2}\left(\alpha+r_0\right).
$$
Obviously, the price increase reads
$$
\frac{3}{2}r_0=\frac{3}{2}\times\$6=\$9.
$$
