A Department store found out that t weeks after the end of a sales promotion the

Question

A Department store found out that t weeks after the end of a sales promotion the volume of Sales was given by S(t) = B + ae^ki (0 lessthanorequalto t lessthanorequalto 4) where B= 50,000 and is equal to the average weekly volume of sales before promotion. the sales volumes at the end of the first and third weeks were $83, 515 and $65, 055, respectively. Assume that the sales volume is decreasing exponentially' Find the decay constant k Find the sales volume at the end of the fourth week. How fast is the sales volume dropping at the end of the fourth week?

Explanation / Answer

given,

S(t)=B+ae-kt

B= 50000

S(t)=50000+ae-kt

given ,

if t = 1 , S(t)= 83515

substituting them gives

83515=50000+ae-k

33515 = ae-k   ................................................. (1)

if t = 3 , S(t)= 65055

substituting them gives

65055=50000+ae-k

15055 = ae-3k   ................................................. (2)

(1)/(2) gives

33515/15055 = ae-k / ae-3k  

2.226 = e2k

2k = ln 2.226

2k = 0.8

k = 0.4

substituting this in (1) gives

33515 = ae-0.4

a = 33515e0.4   

a = 33515*1.4919

a = 50000

so we have, S(t)=50000+50000e-0.4t

b)

in week 4 , t=4

so , S(t)=50000+50000e-0.4*4

=50000+50000e-1.6

= 50000+50000*0.2019

   = 50000+10095

= 60095

c)

S(t)=50000+50000e-0.4t

differentiating with respect to t

dS(t)/dt = 0 + 50000*(-0.4)e-0.4t

dS(t)/dt = -20000e-0.4t

at the end of fourth week , t = 4

dS(t)/dt = -20000e-0.4*4

dS(t)/dt = -20000e-1.6

dS(t)/dt = -20000*0.2019

dS(t)/dt = -4038 sales are dropping so negative

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