Promoters of a major college basketball tournament estimate that the
demand for tickets for *adults* and by *students* are
given by:

\[\begin{aligned} q_a&=5,000-10p_a\\ q_s&=10,000-100p_s\\ \end{aligned}\]

where \(a\) represents adults and \(s\) represents students. They estimate that the marginal and average total cost of seating an additional spectator is constant at $10.

Take each market segment’s demand function that we are given, and solve for the respective price \(p\) to get the two segments’ inverse demand functions. Once we have the inverse demand function, we simply double the slope to find the marginal revenue functions:

\[\begin{align*} q_a&=5,000-10p_a\\ q_a+10p_a&=5,000\\ 10p_a&=5,000-q_a\\ p_a&=500-0.1q_a\\ \end{align*}\]

With this inverse demand function, we double the slope to obtain marginal revenue:

\[MR(q_a)=500-0.2q_a\]

\[\begin{align*} q_s&=10,000-100p_s\\ q_s+100p_s&=10,000\\ 100p_s&=10,000-q_s\\ p_s&=100-0.01q_s\\ \end{align*}\]

With this inverse demand function, we double the slope to obtain marginal revenue:

\[MR(q_s)=500-0.02q_s\]

For each segment, set \(MR=MC\) to obtain the profit-maximizing quantity of tickets to sell. We know the marginal cost for each segment is simply $10. Then, since the firm has market power, raise the price to the maximum each segment is willing to pay (it’s demand at that quantity).

\[\begin{align*} MR(q_a)=MC(q_a)\\ 500-0.2q_a&=10\\ 500&=10+0.2q_a\\ 490&=0.2q_a\\ 2,450&=q_a^{\star}\\ \end{align*}\]

Plug this into the adults’ inverse demand curve to obtain the price:

\[\begin{align*} p_a&=500-0.1q_a\\ p_a&=500-0.1(2,450)\\ p_a&=500-245\\ p_a^{\star}&=\$255\\ \end{align*}\]

\[\begin{align*} MR(q_s)=MC(q_s)\\ 100-0.02q_s&=10\\ 100&=10+0.02q_s\\ 90&=0.02q_s\\ 4,500&=q_s^{\star}\\ \end{align*}\]

Plug this into the students’ inverse demand curve to obtain the price:

\[\begin{align*} p_s&=100-0.01q_s\\ p_s&=100-0.01(4,500)\\ p_a&=100-45\\ p_a^{\star}&=\$55\\ \end{align*}\]

Profit is always price minus average cost (which here is the same as marginal cost, always $10) times quantity.

\[\begin{align*} \pi_a&=(p_a-AC(q_a))q_a\\ \pi_a&=(255-10)2,450\\ \pi_a&=\$600,250\\ \end{align*}\]

\[\begin{align*} \pi_s&=(p_s-AC(q_s))q_s\\ \pi_s&=(55-10)4,500\\ \pi_s&=\$202,500\\ \end{align*}\]

\[\begin{align*} \Pi&=\pi_a+\pi_s\\ \Pi&=$600,250+\$202,500\\ \Pi&=$600,250+\$202,500\\ \Pi&=802,750\\ \end{align*}\]

I did not ask this, but we can easily calculate the markup and the price elasticity of demand for each segment, to show you how the optimal pricing rules remain consistent.

\[\begin{align*} L&=\frac{p-MC(q)}{p}\\ L&=\frac{255-10}{255}\\ L&=0.96\\ \end{align*}\]

About 96% of the price to adults is markup above marginal cost!

The price elasticity of demand for adults at the profit-maximizing price is:

\[\begin{align*} L&=\frac{1}{\epsilon}\\ 0.96&=\frac{1}{\epsilon}\\ \epsilon &= -1.04\\ \end{align*}\]

Adult demand is slightly elastic.

\[\begin{align*} L&=\frac{p-MC(q)}{p}\\ L&=\frac{55-10}{55}\\ L&=0.82\\ \end{align*}\]

About 82% of the price to students is markup above marginal cost.

The price elasticity of demand for students at the profit-maximizing price is:

\[\begin{align*} L&=\frac{1}{\epsilon}\\ 0.82&=\frac{1}{\epsilon}\\ \epsilon &= -1.22\\ \end{align*}\]

Student demand is more elastic.

Hence, we can see because the adults have less elastic demand, the firm will raise the price on them, and they will face a higher markup. Students have more elastic demand, so the firm will lower the price on them (compared to the single price, in question 2, below), and they will face less of a markup.

Simply add the two segments’ demands together to obtain the total market demand:

\[\begin{align*} Q&=Q_a+q_s\\ Q&=(5,000-10p)+(10,000-100p)\\ Q&=15,000-110p\\ \end{align*}\]

Take the market demand and solve for \(p\) to find the inverse demand:

\[\begin{align*} Q&=15,000-110p\\ Q+110p&=15,000\\ 110p&=15,000-Q\\ p&=136.36-\frac{1}{110}Q\\ \end{align*}\]

Note if you round (for example, the slope to 0.009), we may get slightly different answers from here on out.

Knowing the inverse demand function, we can double the slope to obtain the marginal revenue function:

\[MR(Q)=136.36-\frac{2}{110}Q\]

Set \(MR(Q)=MC\) to find the profit-maximizing quantity, then plug this into inverse market demand to find the profit-maximizing price.

\[\begin{align*}
MR(Q)=MC(Q)\\
136.36-\frac{2}{110}Q&=10\\
136.36&=10+\frac{2}{110}Q\\
126.36&=\frac{2}{110}Q\\
6,950&\approx Q^{\star}\\
\end{align*}\]^{1}

Plug this into the market inverse demand curve to obtain the price:

\[\begin{align*} p&=136.36-\frac{1}{110}Q\\ p&=136.36-\frac{1}{110}(6,950)\\ p&=136.36-63.18\\ p^{\star}&=\$73.18\\ \end{align*}\]

Compare this single price to the price charged to the different
segments above. The firm *raised* the price on adults, since they
have less elastic demand, and *lowered* the price on students,
since they ahve more elastic demand.

\[\begin{align*} \pi&=(p-AC(Q))Q\\ \pi_s&=(73.18-10)6,950\\ \pi_s&=\$439,101\\ \end{align*}\]

Note this is smaller than the total profit under price discrimination ($802,750) in part 2c.