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PLEASE HURRY AND ANSWER THIS CORRECTLY USING : Exponential Functions

The function p(w)=240(1.1)^w represents the number of specialty items produced at the old factory w weeks after a change in management. The graph represents the number of specialty items produced at the new factory during the same time period.

PLEASE HURRY AND ANSWER THIS CORRECTLY USING : Exponential Functions The function-example-1
User Zeph
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2 Answers

3 votes

Answer:

(A) 50

(B) Old-10% and New-15%

(C) 6 Weeks

User Josh Matthews
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(a) During Week 0, how many more specialty items were produced at the old factory than at the new factory? Explain.

Old: p(0) = 240 (1.1)^0 = 240

New: (0,190)

That's 50 more items from the old factory in week 0.

Answer: 50

(b) Find and compare the growth rates in the weekly number of specialty items produced at each factory. Show your work.

The growth rate of the old factory is 1.1, 10% per week.

Let's say the new factor is exponential as well. Matching week 0 we get

q(w) = 190 r^w

ln q = ln 190 + w ln r

w ln r = ln q - ln 190

ln r = (ln q - ln 190)/w

r = e^{(ln q - ln 190)/w}

Let's calculate this for a few different points,

r(1) = e^{(ln 220 - ln 190)/1} = 1.1578947368421058

r(2) = e^{(ln 252 - ln 190)/2} = 1.1516578439248717

r(4) = e^{(ln 337 - ln 190)/4} = 1.1540352511556147

r(7) = e^{(ln 505 - ln 190)/7} = 1.14986870386165

That's pretty consistent. We'll go with r=1.15, 15% increase per per week.

Answer: Old factory 10% growth per week, new factor 15% growth per week

(c) When does the weekly number of specialty items produced at the new factory exceed the weekly number of specialty items produced at the old factory? Explain.

We have p(w) = 240(1.1)^2 and q(w) = 190(1.15)^w

240(1.1)^w = 190(1.15)^w

(240/190) (1.1)^w =(1.15)^w

ln(240/190) + w ln (1.1) =w ln(1.15)

ln(240/190) = w ( ln(1.15) - ln (1.1) )

w = ln(240/190) / ln(1.15/1.1) = 5.255468797423731

Answer: By week 6 the new factory exceeds the output of the old

User Tarun Dholakiya
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