In October 2005, a family's gasoline consumption is 35 gallons per month and is increasing at a rate of 2 gallons per month.
The price of gasoline is 250 cents per gallon and is increasing at a rate of 8 cents per month.
Calculate the rate at which the family's monthly gasoline bill is increasing at this time. Give your answer correct to the nearest cent per month.
Rate of increase = ?
I have no idea how to do this!! :(
2006-11-02 15:17:04 · 3 answers · asked by flaquita 1 in Science & Mathematics ➔ Mathematics
Let m = number of months. Then the total consumption is 2m + 35. The price of gasoline is 8m + 250. The total cost at any month is the product of the two, or (2m + 35)(8m + 250), which comes to 16m^2 + 780m + 8750. The equation for the rate increase is found by finding the differential of this quadratic with respect to m. That is, d(16m^2 +780m + 8750)/dm = 32m + 780, which is your answer. At the beginning, where m = 0, the rate of increase is 780 cents/month.
2006-11-02 15:26:04 · answer #1 · answered by Scythian1950 7 · 0⤊ 0⤋
Consumption = 2m + 35 (gallons)
PricePerGallon = 8m + 250 (cents/gallon
Payment = Consumption * PricePerGallon (cents)
Payment = (2m+35)(8m+250) = 16m2 + 280m + 500m + 8750
Payment(m) = 16m2 + 780m + 8750
Rate of increase = d(payment)/dm
Rate of increase(m) = 32m + 780 (cents/month)
October 2005 is m=0 i.e.
The rate of increase is 780 cents/month
2006-11-02 23:39:40 · answer #2 · answered by Sam K 1 · 0⤊ 0⤋
Consumption = 2m + 35 (gallons)
PricePerGallon = 8m + 250 (cents/gallon
Payment = Consumption * PricePerGallon (cents)
Payment = (2m+35)(8m+250) = 16m2 + 280m + 500m + 8750
Payment(m) = 16m2 + 780m + 8750
Rate of increase = d(payment)/dm
Rate of increase(m) = 32m + 780 (cents/month)
October 2005 is m=0
2006-11-02 23:25:18 · answer #3 · answered by Dr. J. 6 · 0⤊ 0⤋