Question

Each firm produces both goods, i.e., good 1 and good 2. Each firm takes the market prices p 0 and p2 2 0 as firm produces T units of good 1 and x2 units of good 2, with (xi, x2) the total costs of C(x.x) = 2i+0.5«% given and chooses output to maximize profits.1 If a R2, it has 1200 (a) (1 point ) For given prices p1 and p2, find the revenue, R(x1, x2), of a single firm (b) (1 points) Find the profit function, II(1,2), of a single firm (c) (10 points) Solve the profit-maximization problem, that is, derive the system of first-order conditions and solve this system. That is, find the critical point of II. Your solution might depend on pı and on p2.

Answer 1

Solution

Back-up Theory

To maximise a function y = f(x), .............................................................................................................................................. (1)

Step 1: Find dy/dx

Step 2: Equate dy/dx to zero and solve for x.

Step 3: Find d²y/dx²

Step 4: Find the value of d²y/dx² at each of the solutions of x obtained in Step 2.

Step 5: Find the solutions of x obtained in Step 2 for which the value of d²y/dx² is negative. These are the values of x at which the function f(x) is maximum. Substitute these values of x in f(x) to get the maximum value(s) of f(x).

If y = f(x₁, x₂), replace dy/dx by ∂y/∂x₁ and d²y/dx² by ∂²y/∂x₁² and proceed as above. Similarly for x₂. .............................. (2)

Now, to work out the solution, ∂π

Part (a)

Revenue function = R(x₁, x₂) = p₁.x₁ + p₂.x₂Answer 1 ...................................................................................................... (3)

Part (b)

Profit = Revenue – Cost.

So, Profit function = π(x₁, x₂) = R(x₁, x₂) - C(x₁, x₂)

= (p₁.x₁ + p₂.x₂) – {(x₁²+ 0.5x₂²)/1200}

= (1/1200)(1200p₁.x₁ + 1200p₂.x₂ – x₁²_- 0.5x₂²) Answer 2.............................................................................................. (4)

Part (c)

Vide (1), (2) and (4),

∂π/∂x₁ = (1/1200)(1200p₁ – 2x₁)

Equating to zero and solving for x₁,

x₁ = 600p₁

∂²π/∂x₁² = - 1/600 < 0 => π is maximum at x₁ = 600p₁....................................................................................................... (5)

Similarly,

∂π/∂x₂ = (1/1200)(1200p₂ – x₂)

Equating to zero and solving for x₂,

x₂ = 1200p₂

∂²π/∂x₁² = - 1/1200 < 0 => π is maximum at x₂ = 1200p₂ .................................................................................................. (6)

(5) and (6) =>

Critical point of π is (600p₁, 1200p₂) Answer 3

DONE