The sales decay for a product is given by $S = 80000e^{-0.9x}$, where $S$ is the monthly sales and $x$ is the number of months that have passed since the end of a promotional campaign. (a) What will be the sales 2 months after the end of the campaign? Round the answer to two decimal places. (b) How many months after the end of the campaign will sales drop below $1000, if no new campaign is initiated? Round up to the nearest whole number.

1. Problem Description

The sales decay for a product is given by $S = 80000e^{-0.9x}$, where $S$ is the monthly sales and $x$ is the number of months that have passed since the end of a promotional campaign.

(a) What will be the sales 2 months after the end of the campaign? Round the answer to two decimal places.

(b) How many months after the end of the campaign will sales drop below $1000, if no new campaign is initiated? Round up to the nearest whole number.

2. Solution Steps

(a) To find the sales 2 months after the end of the campaign, substitute $x = 2$ into the equation for $S$.

$S = 80000e^{-0.9x}$

$S = 80000e^{-0.9(2)}$

$S = 80000e^{-1.8}$

$S = 80000(0.165298888)$

$S = 13223.91104$

Rounding to two decimal places, $S = 13223.91$.

(b) To find how many months after the end of the campaign will sales drop below $1000, we set$S < 1000$and solve for$x$.

$1000 > 80000e^{-0.9x}$

Divide both sides by 80000:

$\frac{1000}{80000} > e^{-0.9x}$

$\frac{1}{80} > e^{-0.9x}$

Take the natural logarithm of both sides:

$ln(\frac{1}{80}) > ln(e^{-0.9x})$

$ln(\frac{1}{80}) > -0.9x$

$-4.38202663466 > -0.9x$

Divide both sides by -0.

9. Since we are dividing by a negative number, flip the inequality sign:

$\frac{-4.38202663466}{-0.9} < x$

$4.86891848295 < x$

$x > 4.86891848295$

Round up to the nearest whole number, $x = 5$.

3. Final Answer

(a)

$13223.91$

(b)

$5$