Finding maximum revenue and price of a revenue function

Question

The John Deere Company has found that the revenue from sales of heavy-duty tractors is a function of the unit price p that it charges. If the revenue $R$ is $R(p) = – \frac{1}{2}p^2 + 1900p$, what unit price p should be charged to maximize revenue? What is the maximum revenue? What recommendations would you give the John Deere Company?

Answer 1

HINT: There are a number of ways to determine the maximum revenue.

1. Graph the function and find the function's maximum.
2. Find the vertex of the parabola using the formula for the $x$-coordinate of the vertex.
3. Take the first derivative of the function and set it to zero and solve.

Which of these have you tried? Which of these do you not understand?

Answer 2

Start by finding the derivative of $R$ with respect to $p$: $$R'(p)=-p+1900$$ Then, to find the maximum, set $R'$ to zero and solve for $p$: $$-p+1900=0$$ $$p=1900$$ So the revenue is maximized when $p=1900$, and the maximum revenue will be $R(1900)$.