Found the following problem answered incorrectly here:

Gloria (a.k.a. Glorious), a well-known fast food operator in the University area seeks your expert advice. She has given you the following information: when she charged a price of $210 for a box lunch there were about 1,500 customers per month and a price of $220 only 1,000 customers. Glorious wants to know the following (based on operating cost of $120 per box lunch):

a) What price would maximize profit?

b) What price would maximize revenue?

c) What is the price elasticity at the profit maximizing price?

Thought I’d help out with my own two cents:

## a) What price would maximize profit?

Assuming there is a linear function for quantity \(Q(P)\) where \(P\) is price.

\(Q(P) = a + bP \)

We want to know the intercept a and the slope b.

First we find the slope:

\(Q(210) = 1500 = a + 210b \)

\(a = 1500 – 210b \)

\(Q(220) = 1000 = a + 220b = (1500 – 210b) + 220b = 1500 + 10b \)

\(10b = 1000 – 1500 \)

\(b = -500/10 = -50 \)

Then the intercept:

\(Q(210) = 1500 = a – 210(50) \)

\(a = 1500 + 10500 = 12000 \)

Thus, we have the following quantity function:

\(Q(P) = a + bP = 12000 – 50P \)

Total revenue is described by the following function:

\(R(P) = Q(P)P = 12000P – 50P^2 \)

Total costs are described by the following function:

\(C(P) = 120Q(P) = 1440000 – 6000P \)

The profit function is thus

\(M(P) = R(P) – C(P) = 12000P – 50P^2 – (1440000 – 6000P) \\= -1440000 + 18000P – 50P^2 \)

To maximizing profit we differentiate the profit function and set it to zero.

\(M'(P) = 18000 – 100P = 0 \)

\(P = 180 \)

**The profit maximizing price is thus $180. **

Let’s see how much profit this gives:

\(M(180) = -1440000 + 18000(180) – 50(180)^2 = 180000 \)

## b) What price would maximize revenue?

To maximizing revenue, while ignoring costs, we have to differentiate the revenue function and set it to zero:

\(R'(P) = 12000 – 100P = 0 \)

\(P = 120 \)

**The revenue maximizing price is thus $120. **

Let’s see how much revenue this gives:

\(R(120) = 12000(120) – 50(120)^2 = 720000 \)

How about profit?

\(M(120) = -1440000 + 18000(120) – 50(120)^2 = 0 \)

So, there is no profit in the revenue maximizing scenario.

## c) What is the price elasticity at the profit maximizing price?

Price elasticity is defined by

\(E(P) = Q'(P)\frac{P}{Q(P)} = \frac{-50P}{12000 – 50P} = \frac{P}{P-240} \)

**The price elasticity at maximized profit is **

\(E(180) = \frac{180}{180-240} = -3 \)

The price is elastic. Somewhat less elastic than Coca-Cola.