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 $\frac{\partial q/\partial p}{q/p}$



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.

  • 1
    $\begingroup$ One hint: The price elasticity is constant -3. With this information you can derive the demand function. See here my answer If I´m right it is $q(p)=p^{-3} \Rightarrow p(q)=q^{-1/3}$ $\endgroup$ – callculus Apr 28 '18 at 22:04
  • $\begingroup$ Additionally I think that the two exercises are independent from each other. The demand function of the first exercise has nothing to do with the demand function of the second exercise. $\endgroup$ – callculus Apr 28 '18 at 22:10
  • $\begingroup$ @callculus I calculated price elasticity and I get the demand function $$q=100/3p$$ for p<20. Why you think they are independent? They are the same question parts. And it ask price change. $\endgroup$ – user489188 Apr 29 '18 at 1:47
  • $\begingroup$ "Its demand function has a constant price elasticity of demand whose value is −3" That´s why I think that it cannot be the demand function of the first part. $\endgroup$ – callculus Apr 29 '18 at 12:10

@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 $$ \frac{3}{2}r_0=\frac{3}{2}\times\$6=\$9. $$


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