Answer:
The correct option is a. $61.25.
Explanation:
Note: The correct cost function of the farmer is as follows:
C(Q) = 0.05Q^2 ……………….. (1)
Differentiating equation
MC = C’(Q) = 0.1Q
P = Expected price = (25% * $3) + (50% * $3.50) + (25% * $4) = $3.50 ……. (2)
Since profit is maximized when MC = P, we equate equations (1) and solve for Q which is the expected profit-maximizing quantity as follows:
0.1Q = 3.50
Q = 3.50 / 0.1 = 35
Substituting Q = 35 into equation (1), we have:
C(Q) = 0.05 * 35^2 = $61.25
R(Q) = Maximum expected revenue = P * Q = $3.50 * 350 = $122.50
The farmer's maximum expected profit = R(Q) - C(Q) = $122.50 - $61.25 = $61.25
Therefore, the correct option is a. $61.25.