Question

The revenue from selling q items is R(q)=600q-q^2 , and the total cost is C(q)= 75+6q. Write a function that gives the total profit earned, and find the quantity which maximizes the profit.Profit pi(q)= ?Profit Maximizing Quantity= ?

Answer 1

`\begin{gathered} R(q)=600q-q^2 \\ C(q)=75+6q \end{gathered}`

Total profit earned:

`\begin{gathered} Pi(q)=R(q)-C(q) \\ \\ Pi(q)=600q-q^2-75-6q \\ Pi(q)=-q^2+594q-75 \\ \end{gathered}`

Profit Maximizing Quantity:

1-Determine marginal revenue by taking the derivative of total revenue

`(d)/(dq)R(q)=600-2q`

2-Determine marginal cost by taking the derivative of total cost

`(d)/(dq)C(q)=6`

3-Set marginal revenue equal to marginal cost and solve for q

`\begin{gathered} 600-2q=6 \\ \\ \text{Subtract 600 in both sides of the equation:} \\ 600-600-2q=6-600 \\ -2q=-594 \\ \\ \text{Divide both sides of the equation into -2}\colon \\ (-2)/(-2)q=(-594)/(-2) \\ \\ q=297 \end{gathered}`Then, the profit maximizing quantity is 297