Q. 35 - Precalculus Enhanced with Graphing Utilities

Question

Suppose that the manufacturer of a gas clothes dryer has found that, when the unit price is p dollars, the

revenue R (in dollars) is $R\left(p\right) =-4p^2+4000p$.

(a) At what prices p is revenue zero?

(b) For what range of prices will revenue exceed $800,000?

Step-by-Step Solution

Answer

(a) The revenue is 0 when price is 1000 dollars.

(b) The solution set is $\left\{x:500-100\sqrt{5}<x<500+100\sqrt{5}\right\}$.

Part (a) Step 1. Given Information

The given function is $R\left(p\right) =-4p^2+4000p$.

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

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

$\begin{array}{rcl}0& =& -4p^2+4000p\\ & =& 4p(-p+1000)\end{array}$

• Equate both the factors with 0.

$\begin{array}{rcl}-4p& =& 0\\ p& =& 0\\ -p+1000& =& 0\\ p& =& 1000\\ & & \end{array}$

• So, the revenue is 0 when the price is 1000 dollars.

Part (b) Step 1. Given Information.

• The given function is $R\left(p\right)=-4p^2+4000p$.
• The minimum revenue is $8000,000.

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

• The minimum revenue is $800,000. So, the function is $-4p^2+4000p>800000 \mathrm{or} -4{\mathrm{p}}^2+4000\mathrm{p}-800000>0$.

• $\begin{array}{rcl}-4p^2+4000p-800000& =& 0\\ -4(p^2-1000p+200000)& =& 0\end{array}$
  

The roots of the function by the quadratic formula are $500\pm 100\sqrt{5}$.

• So, the 0- intercepts are $(500+100\sqrt{5},0),(500-100\sqrt{5},0)$.
• The value of $f\left(0\right)=-800000$.
• So, the y- intercept is $(0,-800000)$.
• The vertex of the parabola is at $p=\frac{-b}{2a}=\frac{-4000}{-8}=-500$
• The value of $f\left(500\right)=200000$.
• So, the vertex of the parabola is at $(500,200000)$.

Step 5. Graph the function

• Plot the graph using the obtained intercepts and vertex.

• The function is above the horizontal axis when $500-100\sqrt{5}<x<500+\sqrt{5}$.
• So, the solution set is $\left\{x:500-100\sqrt{5}<x<500+\sqrt{5}\right\}$