Answer:
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.
