suppose that the cost function of the honey farm is C_H(H, A)  H ²/100  1A and the cost function of the apple orchard is C_A(H, A)  A²/100, where H and A are the number of units of honey and apples produced respectively.The price of honey is $2 and the price of apples is $4 per unit.Let A₁ be the output of apples if the firms operate independently, and let A₂ be the output of apples if the firms are operated by a profit-maximizing single owner.

A) A₁  A₂  200
B) A₁  100 and A₂  200
C) A₁  125 and A₂  200.
D) A₁  200 and A₂  250.
E) A₁  100 and A₂  200.

Question

Anjali Jagga · Accepted Answer

Final Answer : D, Explanation :  To find the optimal output levels for each farm, we need to solve for their respective profit-maximizing quantities.For the honey farm, the profit function is:π(H) = (2H)H - C H (H, A)π(H) = 2H^2 -    H  2 /100    AWe take the derivative with respect to H and set it equal to zero to find the maximum:dπ/dH = 4H -     2 /100 = 0H = (    2 /400)For the apple orchard, the profit function is:π(A) = (4A)A - C A (H, A)π(A) = 4A^2 -    A 2 /100We take the derivative with respect to A and set it equal to zero to find the maximum:dπ/dA = 8A -     2 /100 = 0A = (    2 /800)Now, if the farms operate independently, then we have:A 1  = (    2 /800)A 2  = (    2 /400)Substituting the given prices, we get:A 1  = (1/4)  A 2  = (1/2)  Therefore,A 1     = 100A 2     = 200However, if the farms are operated by a single profit-maximizing owner, then they will produce the quantities that maximize the total profits of both farms. We can find this by adding the profit functions of both farms:π(Total) = π(H) + π(A)π(Total) = 2H^2 -    H  2 /100    A + 4A^2 -    A 2 /100Taking the derivatives with respect to H and A and setting them equal to zero, we get:dπ/dH = 4H -     2 /100 - 2A 2 /100 = 0H = (    2 /400 - A 2 /200)dπ/dA = 8A -     2 /100 - 2H 2 /100 = 0A = (    2 /800 - H 2 /400)Substituting the given prices and solving for H and A, we get:H = 200A = 50Therefore,A 1     = 200A 2     = 250

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