Q. 88 - Precalculus Enhanced with Graphing Utilities

Question

The John Deere company has found

that the revenue, in dollars, from sales of riding mowers is a function of the unit price p, in dollars, that it charges. If the revenue R is $R\left(p\right)=- \frac{1}{2} p^2 + 1900p$

what unit price p should be charged to maximize revenue? What is the maximum revenue?

Step-by-Step Solution

Answer

The price of a single unit is $1900

And the total revenue is $1805000

Step 1: Given information

The given equation is $R\left(p\right)=- \frac{1}{2}{p}^2 + 1900p$

Step 2: Find the vertex

The coefficient of the square term is negative hence the parabola opens downwards hence the maximum value is at the vertex

$$\begin{gathered} x=-\frac{b}{2a} \\ x=\frac{1900}{1} \\ x=1900 \end{gathered}$$

The price of unit is $1900

Step 3: To find the revenue

Substitute 1900 for p  in the quadratic equation

$$\begin{gathered} R\left(p\right)=-\frac{1}{2}{\left(1900\right)}^2+4000\left(1900\right) \\ R\left(p\right)=-\frac{1}{2}\left(3610000\right)+7600000 \\ R\left(p\right)=1805000 \end{gathered}$$

Step 4: Conclusion

The price of a single unit is $1900

And the total revenue is $1805000