Chapter 4.6, Problem 45E

To determine

To find: in which year the revenue was $1.5 million.

Answer to Problem 45E

After 3 years and 9 years the revenue was $1.5 million.

Explanation of Solution

Given information:

The 12 years that a grocery store has been open, its annual revenue $R$ can be modeled by the function

  $R=0.0001\left(-t^4+12t^3-77t^2+600t+13650\right)$

Where $t$ is the number of years since the store opened.

Calculation:

Substituting $R=1.5$ , the equation becomes,

  $\begin{array}{l}0.0001\left(-t^4+12t^3-77t^2+600t+136500\right)=1.5\\ 0.0001\left(-t^4+12t^3-77t^2+600t-1350\right)=0\end{array}$

Because negative real roots and imaginary roots doesn’t make sense, so need to check for only positive real root.

By the Descartes’s rule, the equation has four or two positive real root.

Input $Y_1=-X^4+12X^3-77X^2+600X-1350$ using key

  $0.0001\left(-X\wedge 4+12X\wedge 3-77X{x}^2+600X-1350\right)$ .

The display is shown below:

  

Press The GRAPH key.

  

To find the zero, press $2\text{ndTRACE}2\text{ENTER}$ .

  

From the above display, the function has two zero at approximately $t=3$ and $t=9$ .

Therefore, after 3 years and 9 years the revenue was $1.5 million.