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

Quizplus · Accepted Answer

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 - F1F1F1S1F1F1F10 H ²/100 F1F1F1S1F1F1F10 A We take the derivative with respect to H and set it equal to zero to find the maximum: dπ/dH = 4H - F1F1F1S1F1F1F10 ²/100 = 0 H = (F1F1F1S1F1F1F10 ²/400) For the apple orchard, the profit function is: π(A) = (4A)A - C_A(H, A) π(A) = 4A^2 - F1F1F1S1F1F1F10 A²/100 We take the derivative with respect to A and set it equal to zero to find the maximum: dπ/dA = 8A - F1F1F1S1F1F1F10 ²/100 = 0 A = (F1F1F1S1F1F1F10 ²/800) Now, if the farms operate independently, then we have: A₁ = (F1F1F1S1F1F1F10 ²/800) A₂ = (F1F1F1S1F1F1F10 ²/400) Substituting the given prices, we get: A₁ = (1/4)F1F1F1S1F1F1F10 A₂ = (1/2)F1F1F1S1F1F1F10 Therefore, A₁ F1F1F1S1F1F1F10 = 100 A₂ F1F1F1S1F1F1F10 = 200 However, 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 - F1F1F1S1F1F1F10 H ²/100 F1F1F1S1F1F1F10 A + 4A^2 - F1F1F1S1F1F1F10 A²/100 Taking the derivatives with respect to H and A and setting them equal to zero, we get: dπ/dH = 4H - F1F1F1S1F1F1F10 ²/100 - 2A²/100 = 0 H = (F1F1F1S1F1F1F10 ²/400 - A²/200) dπ/dA = 8A - F1F1F1S1F1F1F10 ²/100 - 2H²/100 = 0 A = (F1F1F1S1F1F1F10 ²/800 - H²/400) Substituting the given prices and solving for H and A, we get: H = 200 A = 50 Therefore, A₁ F1F1F1S1F1F1F10 = 200 A₂ F1F1F1S1F1F1F10 = 250

Suppose That the Cost Function of the Honey Farm Is