A manufacturer can produce at most 140 units of a certain product each year. The demand equation for the product is pequalsq squared minus 100 q plus 4800 and the manufacturer's average-cost function is c overbarequalstwo thirds q squared minus 30 q plus StartFraction 15 comma 000 Over q EndFraction . Determine the profit-maximizing output q and the corresponding maximum profit.

As it can sale up to 140 it should sale 140 units

Explanation:

P = Q^2 - 100Q + 4800

C = 2/3Q^2 -30Q + 15,000Q^-1

Profit per unit P - C = (1/3Q^2 -70Q + 4800 - 15000Q^-1)

Total Profit Profit per unit x Q

1/3Q^3 - 70Q^2 + 4800Q - 15000

As this is a third degree equation the profit scalates therefore the company should sale as much as possibley can to make profit

As it can sale up to 140 it should sale 140 units

Also if we try to detemrinate the point at which Marginal revenue matches Marginal cost:

Revenue

P x Q = (Q^2 - 100Q + 4800) x Q = Q^3 - 100Q^2 + 4800Q

marginal Revenue (slope of the revenue funciton)

3Q^2-200Q+4,800

Cost

C x Q = (2/3Q^2 -30Q + 15,000Q^-1)* Q = 2/3Q^3 -30Q^2 + 15,000

Marginal Cost (slope of the cost function)

2Q^2 - 60Q

Marginal Revenue = Marginal Cost

3Q^2-200Q+4,800 = 2Q^2 - 60Q

Q^2 -140Q + 4800 = 0

When solving with the quadratic formula for the root (profit maximization point we notice the isn't a solution thus It cannot be solved which enhance the previous point that there isn't a profit maximization point in this assignment.

are you looking for a specific answer?

answer; post closing trial balance;