Q) ADVERTISING: Everett Dunn\'s advertising firm has been hired to promote a new

Question

Q) ADVERTISING: Everett Dunn's advertising firm has been hired to promote a new television series for 3 weeks before its debut and 2 weeks afterward. After t weeks of the adveristisng campaing, Everett estimates that P(t) percent of the viewing public is aware of the series, where

P(t) = ( (59 t) / ((0.7 t^2)+16) ) + 6

a) what is the average percentage of the viewing public that is aware of the show during the 5 weeks of the advertising campaign?

b) at what ime during the 5 weeks of the advertising campaign should Everett expect the percentage of viewers to be the same as the average percentage found in part (a) ?

Thank you,

Explanation / Answer

(a) The average is the integral A=1/5 int(0<t<5) P(t) dt = 1/5 * int (0<t<5) (59t/(0.7t^2+16) + 6) dt

A = 6+59/5 int 10t /(7t^2+160) dt =

A = 6+59/7 int 14t /(7t^2+160) dt = (which is of the form u'/u so has integral ln(u))

A = 6+ [59/7 ln(7t^2+160) ](0<t<5) = 6+59/7 ln(67/32) = 12.228

So 12.228% of the viewing public is aware of the show ruing the 5 weeks.

(b) We need to solve P(t) = 12.228

<=> ( (59 t) / ((0.7 t^2)+16) ) + 6 = 12.228

<=> -4.36t^2+59t-99.65 = 0

We find by solving this second order equation t=1.97 and t=11.55 approx

So t=2 is the nearest week before the end of the 5th week

Hence we should expect the percentage of viewers to be the same as average at the 2nd week

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