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given l(v)=8.1-.067v

l(v)= miles per gallon, v = velocity

a. if fuel costs 3.80 per gallon, use the formula for l(v) to determine a function, F(v) which gives fuel cost as a function of driving speed, v.

b. Assume you pay your drivers an hourly wage of 22 per hour, use this fact to find a function W(v) which gives wage cost as a function of driving speed.

c. the total cost, C = m(v), is the sum of the functions m(v) = F(v) + W(v). Determine the driving speed, v, between 45 and 65, that minimizes the total cost, use the second derivative test to classify any critical points.

Thank you so much i have had trouble with this for days..

First let's think about this. If our car has a constant gas mileage of 24 miles per gallon and gas prices is $3.80 per gallon, we will be more interested in how much money we are spending per miles. So it should be 3.80/24 dollars per mile! (i.e. 24 miles later, you would have used up 1 gallon of gas which would cost you $3.80.)

a) The gas mileage is l(v) and the cost of gas is $3.80 per gallon, so

F(v) = 3.8/l(v) = 3.8/(8.1-.067v )

And this is measure in dollar per mile.

b) Speed is distance travelled over time. So that the time spent is Distance / speed.

If the driver is paid $22 per hour, we only need to multiply $22 by the he is driving which is

22 * (distance) / speed

But for the sake of measuring W(v) in the same unit as F(v) (which will be added later),

W(v) = 22/ v

measured in dollar per mile.

c) C = F(v) + W(v) = 3.8/(8.1-0.067v ) + 22/v

dC/dv = 3.8(-0.067) / (8.1-0.067v )^2 - 22/v^2 set to zero and solve to get

v^2 / (8.1-0.067v )^2 = 22/(3.8(0.067))

which implies

v/ (8.1-0.067v ) = sqrt (22/(3.8*0.067) )= 9.2957

=> 1.62281 v = 75.2952

=> v = 46.398

to show that this is the one that minimizes C, we find (the second derivative of C w.r.t. v)

C'' = (-2)3.8(-0.067)(-0.067) / (8.1-0.067v )^3 -(-2)(22)/v^3

= - 2(3.8)(0.067)^2/(8.1-0.067v )^3 + 44 / v^3

At v = 46.398, C'' is positive which by the second derivative test, tells us that

v = 46.398 is a relative minimum. Being the only critical point between 45 and 65,

this relative minimum must be an absolute minimum.

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