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

Answer:

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

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

Step-by-step explanation:

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

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

$S(t) = -\frac{81}{2500}\cdot t^{3} + \frac{267}{250}\cdot t^{2} - 11.9\cdot t + C$ (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:

$-\frac{81}{2500} \cdot 9^{3} + \frac{267}{250}\cdot 9^{2}-11.9\cdot (9) + C = 2.9$

$C = 47.112$

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

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:

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

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

$\bar S = \frac{1}{5}\cdot \left[-\frac{81}{10000}\cdot (13^{4}-8^{4}) +\frac{89}{250}\cdot (13^{3}-8^{3}) -\frac{119}{20}\cdot (13^{2}-8^{2}) +47.112\cdot (13-8) \right]$$\bar{S} = \frac{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.