Bob’s Bats produces baseball bats, and has the following costs: \[\begin{aligned} C(q)&=5q^2+720\\ MC(q)&=10q\\ \end{aligned}\]

and faces a market demand for bats: \[q = 120-0.4p\]

where quantity is measured in thousands of bats

1. Write Bob’s Marginal Revenue function.

If we can find the inverse demand (of the form \(p=a+bx\), we can simply double the slope \((b)\) to get marginal revenue. We have the demand function, so solve it for \(p\) to get the inverse:

\[\begin{align*} q &= 120-0.4p\\ q+0.4p&=120\\ 0.4p&=120-q\\ p&=300-2.5q\\ \end{align*}\]

This is the inverse demand function, so marginal revenue is


2. Find the profit-maximizing quantity and price.

First, the quantity, we follow Rule #1 as always: the profit maximizing \(q^*\) is where \(MR(q)=MC(q)\)

\[\begin{align*} MR(q)&= MC(q)\\ 300-5q&=10q\\ 300&=15q\\ 20&=q^\star\\ \end{align*}\]

Now that we know the profit-maximizing quantity, we need to find the maximum price consumers are willing to pay for 20 units. Plug this into the inverse demand function:

\[\begin{align*} p&=300-2.5q\\ p&=300-2.5(20)\\ p&=300-50\\ p^\star&=250\\ \end{align*}\]

3. How much total profit does Bob’s Bats earn? Should Bob stay or exit this industry in the long run?

Total profit again can be found with Rule #2: \(\pi=[p-AC(q)]q\)

We first need to find the Average Cost function from total cost, by dividing it by \(q\):

\[AC(q) = \frac{C(q)}{q} = \frac{5q^2+720}{q} = 5q+\frac{720}{q}\]

Now we specifically need to find the average cost at 20 units:

\[\begin{align*} AC(q)&=5q+\frac{720}{q}\\ AC(20)&=5(20)+\frac{720}{(20)}\\ AC(20)&=100+36\\ AC(20)&=136\\ \end{align*}\]

Now just plug in the price, average cost, and quantity:

\[\begin{align*} \pi&=[p-AC(q)]q\\ \pi&=[250-136]20\\ \pi&=[114]20\\ \pi&=2,280\\ \end{align*}\]

4. At what price would Bob’s Bats break even?

From before, we know that a firm’s break even price is at the minimum of its Average Cost curve, where Average Cost is equal to Marginal Cost. First let’s find the quantity where that happens:

\[\begin{align*} AC(q)&=MC(q)\\ 5q+\frac{720}{q}&=10q\\ \frac{720}{q}&=5q\\ 720&=5q^2\\ 144&=q^2\\ 12&=q\\ \end{align*}\]

This the quantity where AC is minimized and equal to MC. We need to find the price, so plug this quantity into either AC or MC. MC is easier here:

\[\begin{align*} MC(q)&=10q\\ MC(12)&=10(12)\\ MC(12)&=120\\ \end{align*}\]

The firm breaks even at a price of $120.

5. How much of Bob’s price is markup (over marginal cost)?

Use the Lerner Index: \(L=\frac{p-MC(q)}{p}\). This will tell us what proportion of the price is markup above marginal cost.

First, we do need to find the marginal cost at \(q^*=20\):

\[\begin{align*} MC(q)&=10q\\ MC(20)&=10(20)\\ MC(20)&=200\\ \end{align*}\]

Now plug this and \(p^*\) into the Lerner index:

\[\begin{align*} L &= \frac{p-MC(q)}{p}\\ L &= \frac{250-200}{250}\\ L &=\frac{50}{250}\\ L &= 0.20\\ \end{align*}\]

The Lerner index says that 20% of the firm’s price ($250) is markup above marginal cost ($200).

6. Calculate the price elasticity of demand at Bob’s profit-maximizing price.

While you could calculate this manually, it’s a lot faster to use the full Lerner Index equation: \(L=\frac{p-MC(q)}{p}=-\frac{1}{\epsilon}\). Since we know \(L\), we can set it equal to \(-\frac{1}{\epsilon}\) and solve for \(\epsilon\):

\[\begin{align*} L&=-\frac{1}{\epsilon}\\ 0.20&=-\frac{1}{\epsilon}\\ 0.20\epsilon &=-1\\ \epsilon &=-\frac{1}{0.20}\\ \epsilon&=-5\\ \end{align*}\]

Demand is elastic. For every 1% the price increases (decreases), consumers will buy 5% less (more).