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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..

The trick to this question seems to be to realize that the "cost" really refers to a cost per mile or per time. So for part a, assuming we want $/mile, we have

p = $3.80/gal and I(v) = miles/gal

which need to be combined to give cost/mi = F(v) = $/mi = p/I(v) = ($/gal)/(mi/gal).

Similarly, for the drivers earning q = $22/hour, we get

cost/mi = W(v) = q/v = ($/hr)/(mi/hr) = $/mi.

It is important to note that in order to get the correct dimension for the cost function, i.e., $/mi, I took the 2 quantities I had to work with, namely for part a the price per gallon, $/gal, and miles/gal, and combined them in such a way as to get the proper dimensions. Checking dimensions in a problem like this is a really good habit to get into.

Back to the problem, the sum is thus

m(v) = F(v) + W(v) = 3.8/(8.1 - (0.067)v + 22/v.

You then need to take the derivative of m(v), which is easy, and set it equal to 0. You will get a quadratic equation in v which you can solve using the quadratic equation. Doing this I get a value of v = 46.3 for m(v) a minimum.

Taking the second derivative of m(v) and evaluating it at this value of v you should get a positive result, which means the slope is increasing with v, corresponding to a minimum.

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