The problem describes the sales decay of a product using the formula $S = 80000e^{-0.9x}$, where $S$ is the monthly sales and $x$ is the number of months after the end of a promotional campaign. (a) We need to find the sales 2 months after the end of the campaign, rounded to two decimal places. (b) We need to find how many months it takes for the sales to drop below $1000, rounded up to the nearest whole number.

1. Problem Description

The problem describes the sales decay of a product using the formula $S = 80000e^{-0.9x}$, where $S$ is the monthly sales and $x$ is the number of months after the end of a promotional campaign.

(a) We need to find the sales 2 months after the end of the campaign, rounded to two decimal places.

(b) We need to find how many months it takes for the sales to drop below $1000, rounded up to the nearest whole number.

2. Solution Steps

(a) To find the sales 2 months after the campaign, we substitute $x = 2$ into the formula:

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

$S = 80000e^{-1.8}$

$S \approx 80000(0.1652988882)$

$S \approx 13223.911056$

Rounding to two decimal places, we get $S \approx 13223.91$.

(b) To find the number of months it takes for sales to 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:

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

$ln(1) - ln(80) > -0.9x$

$0 - ln(80) > -0.9x$

$-ln(80) > -0.9x$

Divide both sides by -0.

9. Remember to flip the inequality sign:

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

$\frac{ln(80)}{0.9} < x$

$x > \frac{4.38202663466}{0.9}$

$x > 4.86891848296$

Since we need to round up to the nearest whole number, $x = 5$.

3. Final Answer

(a) $13223.91

(b) 5