## Calculate Equilibrium and Demand Curve

There are n = 25 restaurants in a downtown area open for lunch. Each one has a monopoly on their specific cuisine (there is only one Italian, only one Mexican, only one Japanese, and so on for twenty-five different unique gastronomic alternatives). However, there is high substitutability from a customer's perspective, as diners like a variety depending on taste and, of course, prices. Restaurant I faces the following individual demand: Di(pi) = (1000/n) - pi. The cost function for restaurant I is C(xi) = xi^2 + 50.

- What is the inverse demand curve that restaurant faces? What is the marginal revenue MRI and the marginal cost MCI?
- How many customers xi will the restaurant I serve, and what price pi will it advertise?
- How much profit will the restaurant make? Since all other restaurants have the same demand and costs as restaurant I, is having in this downtown profitable? With low barriers to entry how many new restaurants will open? As n increases, Di(Pi) falls until it is tangent to AC. Find equilibrium n for which profits are zero.
- Starting from the long-run equilibrium n you found in part c, the cost of leases now increases, and the new cost function that each restaurant faces becomes C(Xi)=Xi^2 + 200. What is the new equilibrium number of firms in the market?

**Solution:**

**a)**

Qi=1000/25-pi=40-pi

Pi=40-Qi{ inverse demand}

MRi=40-2qi

MCi=∆TC/∆Q=2qi

**b)**

Firm Profit Maximizing quantity is at MRi=MCi

40-2qi=2qi

40=4qi

Qi=40)4=10

Pi=40-10=30

**c)**

πi = pi*xi – xi ^2 – 50

we have : xi = 10 and pi = 30,

then πi = 150

Yes it’s profitable because πi<0

ACi = C(xi) / xi = xi + 50 /xi

Min (ACi) :

∆ACi/∆xi = 0 ↔ 1 – 50/(xi^2) = 0 ↔ xi = √50

Pi = ACi↔ 1000/n - xi = xi + 50 /xi

↔ 1000/n = 2* xi + 50 /xi = 3*√50

↔n = 1000/ (3*√50)

↔ n = 47

**d)**

ACi = C(xi) / xi = xi + 200 /xi

Min (ACi) :

∆ACi/∆xi = 0 ↔ 1 – 200/(xi^2) = 0 ↔ xi = √200

Pi = ACi↔ 1000/n - xi = xi + 200 /xi

↔ 1000/n = 2* xi + 200 /xi = 3*√200

↔ n = 1000/ (3*√200)

↔ n = 24

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