The management of a chain of frozen yogurt stores believes that t days after the end of an advertising campaign, the rate at which the volume V (in dollars) of sales is changing is approximated by V ' (t) = - 26400 e - 0.49 t. On the day the advertising campaign ends (t = 0), the sales volume is $ 170, 000. Find both V ' (6) and its integral V (6). Round your answers to the nearest cent.

Step-by-step explanation:

Given the rate at which the volume V (in dollars) of sales is changing is approximated by the equation

V ' (t) = - 26400 e^ - 0.49 t.

t = time (in days)

. v' (6) can be derived by simply substituting t = 6 into the modelled equation as shown:

V' (6) = - 26400 e - 0.49 (6)

V' (6) = - 26400e-2.94

V' (6) = - 26400*-0.2217

V' (6) = $5852.88

V' (6) = $5,853 to nearest dollars

V' (6) = 585300cents to nearest cent

To get v (6), we need to get v (t) first by integrating the given function as shown:

V (t) = ∫-26400 e - 0.49 t dt

V (t) = - 26,400e-0.49t/-0.49

V (t) = 53,877.55e-0.49t + C.

When t = 0, V (t) = $170,000

170,000 = 53,877.55e-0 + C

170000 = 53,877.55 (2.7183) + C

170,000 = 146,454.37+C

C = 170,000-146,454.37

C = 23545.64

V (6) = 53,877.55e-0.49 (6) + 23545.64

V (6) = - 11,945.63+23545.64

V (6) = $11,600 (to the nearest dollars)

Since $1 = 100cents

$11,600 = 1,160,000cents