Exponential Function word problem

Question

Elizabeth’s first business started in 2015, sells homemade brownie mixes. The profit during her 15th month of business was 50% higher than in her 3rd month of business. She predicts that her profit is growing exponentially.

(a) If this assumption is true, in what month will her profit from the brownie business be three times what it was in the first month? You can round your answer to the nearest month.

(b) Elizabeth’s second business sells handmade beaded jewellery. She earned a total profit of $258during her 2nd month of business and a total profit of $320 during her 4th month of business.

i. Based on the data for these two months, if the profit for Elizabeth’s jewellery business is modelled by a linear function, what would the model predict her profit during her eighth month of business to be?

ii. Based on the data for these two months, if the profit for Elizabeth’s jewellery business was instead modelled by an exponential function, what would the model predict her profit during her eighth month of business to be?

Answer 1

brownies

M(t)=A*(1.0344)^(t-1) is equation for money where t is in months and A is first month profit

check, M(15)/M(3)=1.0344¹²=1.5 or 50% gain

found 1.0344 by dividing 15th month money by 3rd monh money and taking twelve root

correction, changed 1.3A (30% more) to 3A (three times more).

M(t)=3A ,,,3 times more than first month

when 3A=A*(1.0344)^(t-1)

ln3=(t-1)*ln(1.0344)

t-1=32.48

t=33.65 months

jewelery

62/2=31 for linear need constant additive growth per month of $31. 2 months=62, 1 month=62/2

M=31*t+196,,t is in months,, t=2,M=258,,,t=4,M=320

M(8)=31*8+196 (196 is the result of letting t=0 which is back two months from t=2)

since there are two months need to take square root of 320/258 for base of exponentiation

√(320/258)=1.11

M=A*1.11^(t-1)

when t=8 ,,,M=A*2.07,,,,,,,,,,,,,, 1.11⁷=2.07

meaning money in 8th month 207% higher than first month