SOLUTION: The number of miles, M, that a car can travel on a gallon of gas during a trip depends on the average speed, v, of the car on the trip according to the model M = -.011v^2 + 1.3v.
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-> SOLUTION: The number of miles, M, that a car can travel on a gallon of gas during a trip depends on the average speed, v, of the car on the trip according to the model M = -.011v^2 + 1.3v.
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Question 372800: The number of miles, M, that a car can travel on a gallon of gas during a trip depends on the average speed, v, of the car on the trip according to the model M = -.011v^2 + 1.3v. Find the average speed that will minimize the costs for a trip. A person wants to take a 1900 mile trip in the car and wants to spend less than $55 for gas. If gas costs $1.10, how fast can the person drive? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! The number of miles, M, that a car can travel on a gallon of gas during a trip
depends on the average speed, v, of the car on the trip according to the model
M = -.011v^2 + 1.3v.
Find the average speed that will minimize the costs for a trip.
:
Minimum cost will be at Maximum mpg, which occurs at the axis of symmetry.
Find the axis of symmetry of the equation M = -.011v^2 + 1.3v
v =
v =
v = +59.1 mph gives max mpg (M)
;
;
A person wants to take a 1900 mile trip in the car and wants to spend less than $55 for gas.
If gas costs $1.10, how fast can the person drive?
:
Solve for M
1.10(1900/M) < $55
Divide both sides by 1.10, results < 50
M >
M > 38 mpg (More than 38 mpg to spend under $55)
:
Find the speed (v) which gives > 38 mpg. Let's use 38.1 mph
-.011v^2 + 1.3v = 38.1
-.011v^2 + 1.3v - 38.1 = 0
Solve this using the quadratic formula, a=-.011; b=1.3; c=-38.1
do the math, you will get two solutions of
53.8 mph and 64.4 mph
Perhaps we can say if he drives between these two speeds he will spend <$55
