# Monopoly with price discrimination

A city has a single electricity supplier. Electricity production cost is $c$ per unit. Utility function $U(q,t)=a\ln(1+q) - t$, where $q$ is electricity consumption and $t$ is electricity tariff. What is the profit maximizing tariff?

I have figured out $\text{MC}=c$ and the supplier is monopolist so $\text{MR}=\text{MC}$ condition have to be used but I can't figure out how to get MR from the utility function.

• Welcome to Economics SE. In its current form your question will probably be closed. One reason is that we prefer not to have images as the text contained in them does not show up in search engines. The other reason is that this is a homework/exam prep. question with no effort shown on your part. So please take the time to type in the question and tell us if you can identify what model this is. Then we may be able to help you. Jun 29, 2015 at 9:34
• I am reopening the question because the author has edited it in accordance with the suggestions made above. Jun 29, 2015 at 16:48
• Should the utility function be $U(q,t)=a\ln(1+q)-tq$—i.e. is $t$ the tariff per-unit of electricity consumed? Jun 29, 2015 at 16:50
• yes , t is tariff per unit consumed. Jun 29, 2015 at 18:21
• The original question was about second degree price discrimination. I believe the OP inadvertently deleted the part about the two kinds of consumers. @light can you please provide reasoning why MR = MC? I also strongly suspect that it is either $-t \cdot q$ in the utility function or that $t$ is the total amount payed for electricity. Jun 29, 2015 at 21:35

I assume $t=pq$ is meant. $p$ is the price per unit.

We want to know when

$$U(q,p)=a \ln(1+q) - pq$$

is maximal, given $a>0$ and $p>0$, since this is the utility function and the quantity $q$ the customer buys is the quantity where the utility function is maximal.

The derivative with respect to $q$ is

$$\frac{d}{dq} U(q,p) = \frac{a}{1+q} - p$$

Set derivative to zero gives $q =\frac{a-p}{p}$.

Now we want to know when $(p-c)q$ is maximal, since that gives maximal profit.

Insert $q =\frac{a-p}{p}$ to get $\frac{(a-p)(p-c)}{p}$ is maximal. Rewrite this to $-p+a+c-\frac{ac}{p}$. Take the derivative with respect to $p$ and set to zero to get

$$-1+\frac{ac}{p^2}=0$$

$$-p^2+ac=0$$

$$p^2=ac$$

$$p=\sqrt{ac}$$

Thus we get $t=pq=a-p=a-\sqrt{ac}$.