Question

You are given the following C = 1.5Q^2 + 40A + 100 p = 140 - 1Q "where Q is​ output, p is​ price, and C is the total cost of production. Determine the profit-maximizing LOADING... price and output for a monopoly."

Answer 1

q = 20

P = 120

Step-by-step explanation:

`\left \{ {{C = 1.5Q^(2) +40Q + 100} \atop {p = 140 - 1Q}} \right.`

From the Price formula, we deduce that total revenue and from there, the marginal revenue

Total revenue = Price x Q

Total revenue = (140 - 1Q) x Q

Total revenue = 140Q - 1Q^2

Now marginal revenue will be dTR/dQ'

marginal revenue = 140 - 2Q

Now we solve for marginal cost:

Total Cost 1.5Q^2 +40Q + 100

We calculate dTC/dQ'

Marginal Cost = 3Q + 40

Now we do MR = MC

140 - 2Q = 3Q + 40

100 = 5Q

Q = 20

Price= 140 - 1Q = 140 - 1 x (20) = 120

The profit-maximizing Q is 20 and P equal to 120

We can try to check:

Total Revenue = $121 x 19Q = 2,299

Total Revenue = $120 x 20Q = 2,400

Total Revenue = $119 x 21Q = 2,499

Total Cost 1.5Q^2 +40Q + 100

Q = 19 TC = 1401.5

Q = 20 TC = 1500

Q = 21 TC = 1601.5

Profit:

at Q 19 2,299 - 1,401.5 = 897.5

at Q 20 2,400 -1,500 = 900

at Q 21 2,499 - 1,601.5 = 897,5‬

At Q = 20 the profit is maximize From there, adding a new unit decrease the profti for the company.