Question

The rate of change in sales for Garmin from 2008 through 2013 can be modeled by

dS/dT = -0.0972t^2 + 2.136t -11.9
where S is the sales (in billions of dollars) and is the time in years, with t = 8 corresponding to 2008. In 2009, the sales for Garmin were $2.9 billion.
A. Find the model for the sales of Garmin.
B. What were the average sales of Garmin from 2008 through 2013?

Answer 1

a) The model for the sales of Garmin is represented by
`S(t) = -(81)/(2500)\cdot t^(3) + (267)/(250)\cdot t^(2) - 11.9\cdot t + 47.112`.

b) The average sales of Garmin from 2008 through 2013 were $ 2.5 billion.

Explanation:

a) The model for the sales of Garmin is obtained by integration:

`S(t) = -0.0972\int {t^(2)} \, dt + 2.136\int {t}\,dt -11.9 \int\,dt`

`S(t) = -(81)/(2500)\cdot t^(3) + (267)/(250)\cdot t^(2) - 11.9\cdot t + C` (1)

Where
`C` is the integration constant.

If we know that
`t = 9` and
`S(t) = 2.9`, then the model for the sales of Garmin is:

`-(81)/(2500) \cdot 9^(3) + (267)/(250)\cdot 9^(2)-11.9\cdot (9) + C = 2.9`

`C = 47.112`

The model for the sales of Garmin is represented by
`S(t) = -(81)/(2500)\cdot t^(3) + (267)/(250)\cdot t^(2) - 11.9\cdot t + 47.112`.

b) The average sales of the Garmin from 2008 through 2013 (
`\bar{S}`) is determined by the integral form of the definition of average, this is:

`\bar{S} = (1)/(13 - 8) \cdot \int\limits^(13)_(8) {S(t)} \, dt` (2)

`\bar S = (1)/(5)\cdot \int\limits^(13)_(8) {\left[-(81)/(2500)\cdot t^(3) + (267)/(250)\cdot t^(2)-11.9\cdot t + 47.112 \right]} \, dt`

`\bar S = (1)/(5)\cdot \left[-(81)/(10000)\cdot (13^(4)-8^(4)) +(89)/(250)\cdot (13^(3)-8^(3)) -(119)/(20)\cdot (13^(2)-8^(2)) +47.112\cdot (13-8) \right]`
`\bar{S} = (1)/(5)\cdot (-198.167+599.86-624.75+235.56)`

`\bar{S} = 2.5`

The average sales of Garmin from 2008 through 2013 were $ 2.5 billion.