Finding maximum revenue and price of a revenue function
By Michael Henderson •
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?
$\endgroup$2 Answers
$\begingroup$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 $R(1900)$.
$\endgroup$ $\begingroup$HINT: There are a number of ways to determine the maximum revenue.
- Graph the function and find the function's maximum.
- Find the vertex of the parabola using the formula for the $x$-coordinate of the vertex.
- 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?
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