For each carload of ore removed from the Matchless Mine, 1 pound of lead and 1 ounce of silver is smelted. The cost of mining and processing 1 carload of ore is C(Q) = 300 + 3Q + 0.02Q². The demand for silver (per ounce) is P_s = 30 - 0.10Q_s and the demand for lead (per pound) is P_l = 6 - 0.03Q_l. What should the owners of the Matchless Mine do to maximize profits (in the long run)?

Question

Quizplus · Accepted Answer

To maximize profits, we need to find the quantity of ore to mine and process that will yield the highest profit. We can find this by setting the marginal revenue equal to the marginal cost and solving for the quantity. In other words, we want to find the quantity of ore that will maximize the difference between total revenue and total cost. Let's start by calculating the revenue from selling silver and lead. For each carload of ore, we get 1 pound of lead and 1 ounce of silver. Using the demand functions, we can find the price that each metal will fetch: P_s = 30 - 0.10Q_s P_l = 6 - 0.03Q_l For a given quantity of silver or lead, we can find the total revenue by multiplying the quantity by the price. For example, the total revenue from selling Q ounces of silver and P pounds of lead is: TR(Q,P) = Q * (30 - 0.10Q) + P * (6 - 0.03P) The cost of mining and processing 1 carload of ore is C(Q) = 300 + 3Q + 0.02Q². The total cost of mining and processing Q carloads of ore is: TC(Q) = 300Q + 3Q^2 + 0.02Q^2(1/2) = 300Q + 3Q^2 + 0.02Q^(3/2) Now we can find the profit function P(Q,P): P(Q,P) = TR(Q,P) - TC(Q) To find the maximum profit, we need to find the values of Q and P that maximize P(Q,P). We can do this by taking partial derivatives of P with respect to Q and P, and setting them equal to zero: ∂P/∂Q = 30 - 0.2Q - 300 - 6Q = 0 ∂P/∂P = 6 - 0.06P - 300 - 2P = 0 Solving for Q and P, we get: Q = 112.5 P = 100 This means that the owners of Matchless Mine should mine and process 112.5 carloads of ore, and sell 112.5 ounces of silver and 100 pounds of lead to maximize their profits. Therefore, the answer is D.

For Each Carload of Ore Removed from the Matchless Mine