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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= ?

Respuesta :

[tex]\begin{gathered} R(q)=600q-q^2 \\ C(q)=75+6q \end{gathered}[/tex]

Total profit earned:

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

Profit Maximizing Quantity:

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

[tex]\frac{d}{dq}R(q)=600-2q[/tex]

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

[tex]\frac{d}{dq}C(q)=6[/tex]

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

[tex]\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 \\ \frac{-2}{-2}q=\frac{-594}{-2} \\ \\ q=297 \end{gathered}[/tex]Then, the profit maximizing quantity is 297

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