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Calculus/Max/ min fuel costs help :(


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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