Question

sales of a new model of digital camera are approximated by ( s(x)=5400 - 3900e^{-x} ), where ( x ) represents the number of years that the digital camera has been on the market, and ( s(x) ) represents sales in thousands of dollars.
(a) find the sales in year 0.
(b) when will sales reach ( $4,873,000 )?
(c) find the limit on sales.
(a) the sales in year 0 were ( $square ).
(b) the sales will reach ( $4,873,000 ) in about ( square ) years.
(c) the limit on sales is ( $square ).

Explanation:

Step1: Calculate sales in year 0

Substitute \(x = 0\) into \(S(x)=5400 - 3900e^{-x}\).
Since \(e^{0}=1\), then \(S(0)=5400-3900\times1\).
\(S(0)=5400 - 3900=1500\) (in thousands of dollars).

Step2: Find when sales reach \(4873\) (in thousands of dollars)

Set \(S(x)=4873\), so \(4873 = 5400-3900e^{-x}\).
First, rearrange the equation: \(3900e^{-x}=5400 - 4873\).
\(3900e^{-x}=527\).
Then \(e^{-x}=\frac{527}{3900}\).
Take the natural - logarithm of both sides: \(-x=\ln(\frac{527}{3900})\).
\(x=-\ln(\frac{527}{3900})\approx3\).

Step3: Find the limit on sales

We know that \(\lim_{x
ightarrow+\infty}e^{-x}=0\).
For \(S(x)=5400 - 3900e^{-x}\), \(\lim_{x
ightarrow+\infty}S(x)=\lim_{x
ightarrow+\infty}(5400 - 3900e^{-x})\).
Using the limit rule \(\lim_{x
ightarrow+\infty}(a + b f(x))=a + b\lim_{x
ightarrow+\infty}f(x)\) (where \(a = 5400\), \(b=-3900\), \(f(x)=e^{-x}\)), we get \(\lim_{x
ightarrow+\infty}S(x)=5400-3900\times0 = 5400\) (in thousands of dollars).

Answer:

(a) The sales in year 0 were \(\$1500000\).
(b) The sales will reach \(\$4873000\) in about \(3\) years.
(c) The limit on sales is \(\$5400000\).