Question

In the market for lock washers, a perfectly competitive market, the current equilibrium price is $5 per box. Washer King, one of the many producers of washers, has a daily short-run total cost given by TC = 190 + 0.20Q + 0.0025Q2, where Q measures boxes of washers. Washer King's corresponding marginal cost is MC = 0.20 + 0.005Q. How many boxes of washers should Washer King produce per day to maximize profit?

Answer 1

To maximize profit, Washer King should produce approximately 960 boxes of washers per day.

Step-by-step explanation:

To maximize profit, Washer King should produce a quantity of washers where the marginal cost (MC) is equal to the market price. In this case, the equilibrium price is $5 per box. Washer King's marginal cost is given by MC = 0.20 + 0.005Q, where Q represents the number of boxes of washers produced per day. Setting MC equal to $5, we can solve for the quantity of washers to be produced.

0.20 + 0.005Q = 5

0.005Q = 4.80

Q = 960

Therefore, Washer King should produce approximately 960 boxes of washers per day to maximize profit.

Answer 2

The number of boxes of washers Washer King should produce per day to maximize profit = 960 boxes.

And the corresponding maximum daily profit = $2,114

Step-by-step explanation:

The daily, short-run total cost of producing Q boxes of the product is given as

TC = 190 + 0.20Q + 0.0025Q²

The unit price of the product = $5.

Total revenue = (Unit Price) × (Quantity sold) = 5Q

Profit = (Revenue) - (Total Cost)

Profit = 5Q - (190 + 0.20Q + 0.0025Q²)

Profit = P(Q) = -190 + 4.8Q - 0.0025Q²

To maximize the profits, we just obtain the point where the profit function reaches a Maximum.

At the maximum of a function, (dP/dQ) = 0 and (d²P/dQ²) < 0

Profit = P(Q) = -190 + 4.8Q - 0.0025Q²

(dP/dQ) = 4.8 - 0.005Q

At maximum point,

(dP/dQ) = 4.8 - 0.005Q = 0

Q = (4.8/0.005) = 960 boxes

(d²P/dQ²) = -0.005 < 0 (hence, showing that the this point corresponds to a maximum point truly)

Hence, the number of boxes of washers Washer King should produce per day to maximize profit = 960 boxes.

The corresponding maximum profit is then obtained from

P(960) = -190 + (4.8×960) - 0.0025(960²)

Maximum daily profit = $2,114

Hope this Helps!!!