This is a continuation of another question I am looking to f

This is a continuation of another question. I am looking to find the answers for #5 and #6. The questions & answers to #1-4 are posted below for reference. Thank you.

After an extensive study, the Interstate Commerce Commission has concluded that, though maintenance and replacement costs for vehicles are considerable, they do not vary significantly from carrier to carrier. So I won\'t consider vehicle maintenance and repair when trying to optimize my profits.

Solution

5. At optimal speed, W/S = C/( 11- 0.1*S – 0.00002*L) Therefore, W/55 = C/ ( 11 – 0.1*55- 0.00002*75000) = C/ ( 11- 5.5 – 1.5) = C/4 = F(as F = C/No. of miles per gallon) or, W = 55F

6. If W = 55F and if L is to be lower than 75000lbs, then W/S = C/( 11 – 0.1*55- 0.00002*L) or,W/S = C/( 9.5 – 0.00002L) or, S = (W/C)* ( 9.5 – 0.00002L) = (55F/C)*( 9.5 – 0.00002L) = [55/( No of miles per gallon)]* ( 9.5 – 0.00002L). Now, C = fuel cost per gallon and F = fuel cost per mile so that C/F = number of miles per gallon. Now, if L decreases, then 9.5- 0.00002L will increase so that S is expected to increase. Again, if S increases, then the number of miles per gallon decreases so that 55/( number of miles per gallon) increases, which , in turn, increases S. Thus S or, the optimal speed will increase.