Howard J. answered • 12/04/19
Principal Mechanical Engineer with >30 years' math coaching experience
The price of a gallon of milk has been rising about 1.36% per year since 2000.
A.If milk costs $4.70 now, what will it cost next year?
B.If milk costs $4.70 now, how long will it take for the price to top $5?
What type of function did you use to model the scenario? Be specific.
Why did you choose that type of function? What characteristics of the model led you to make that choice?
Did you write a formula for the function? If not, do so now. Be sure to state what the independent and dependent variable represent in the model.
The price is compounding annually like simple interest does:
F=P(1+I/k)n
where I=the annual rate, k=the number of the number of compounding periods per year=1, n=the number of compounding periods beyond the year 2000, P=price in the year 2000, and F=the price in the future treating 2000 as the present.
F=P(1+0.00136)n=1.00136nP
Given F=4.70 when n=19,
$4.70=1.0013619P
P=$4.70(1.00136)-19
=$4.58=the price in the year 2000
so F=4.58(1.00136)n
Let's check the formula for n=19 just to be sure: F=$4.58(1.00136)19=$4.70
A. When the price is $4.70, what's the price the following year?
The following year n=20 so F=4.58(1.00136)20=$4.71
Another way: Let P=the price this year=$4.70 and set n=1
F=4.70(1.00136)1=$4.71
B. When the price tops $5.00, we know F=5.00 and P=4.70 but n=?
5=4.7(1.00136)n
ln(5)=ln[4.7(1.00136)n]
ln(5)=ln(4.7)+ln(1.00136)n
ln(5)-ln(4.7)=nln(1.00136)
ln(5/4.7)=nln(1.00136)
n=ln(5/4.7)/ln(1.00136)=45.5
n=46 years to top $5.00
Check:
F=4.7(1.00136)46
=$5.00 OK
Doubtful the price will outpace inflation!
