Q. 36 - Precalculus Enhanced with Graphing Utilities

Question

The John Deere company has found that the revenue from sales of heavy-duty tractors is a function of the unit price p, in dollars, that it charges. If the revenue R, in dollars, is $R\left(p\right)=-\frac{1}{2}{p}^2+1900p$

(a) At what prices p is revenue zero?

(b) For what range of prices will revenue exceed $1,200,000?

Step-by-Step Solution

Answer

• The revenue is 0 when the price is 3800 dollars.
• The solution set is $\left\{x| 800<x<3000\right\}$.

Part (a) Step 1. Given Information

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

Part (a) Step 2. Find p when revenue is 0

• Substitute 0 for $R\left(p\right)$ into the given function and then factorize.

$\begin{array}{rcl}0& =& -\frac{1}{2}{p}^2+1900p\\ -\frac{1}{2}p(p-3800)& =& 0\end{array}$

• Equate both the factors with 0.

$\begin{array}{rcl}-\frac{1}{2}p& =& 0\\ p& =& 0\\ p-3800& =& 0\\ p& =& 3800\end{array}$

• So, the revenue is 0 when the price is 3800 dollars.

Part (b) Step 1. given Information:

• The given equation is $R\left(p\right)=-\frac{1}{2}{p}^2+1900p$.
• The minimum revenue is $1,200,000.

Part (b) Step 2. Create a function and find intercepts

• The minimum revenue is $800,000. So, the function is $-\frac{1}{2}{p}^2+1900p>1200000 \mathrm{or} -\frac{1}{2}{\mathrm{p}}^2+1900\mathrm{p}-1200000>0$.

• Equate the obtained expression on the left-hand side of the inequality with 0 and find the roots.

  
  

$\begin{array}{rcl}-\frac{1}{2}{p}^2+1900p-1200000& =& 0\\ p^2-3800p+2400000& =& 0\\ (p-3000)(p-800)& =& 0\end{array}$

• So, the x- intercepts are $(800,0),(3000,0)$.
• The value of $f\left(0\right)=-1200000$.
• So, the y- intercept is $(0,-1200000)$.
• The vertex of the parabola is at $x=\frac{-b}{2a}=\frac{-1900}{-1}=1900$
• The value of $f\left(1900\right)=605000$.
• So, the vertex of the parabola is at $(1900,605000)$.

Step 5. Graph the function

• Plot the graph using the obtained intercepts and vertex.

• The function is above the horizontal axis when $800<x<3000$.
• So, the solution set is $\left\{x:800<x<3000\right\}$.