A truck has a minimum speed of 11 mph in high gear. When traveling x mph, the tr

Question

A truck has a minimum speed of 11 mph in high gear. When traveling x mph, the truck burns diesel fuel at the rate of 0.0054796(676/x+x) gal/mile. Assuming that the truck can not be driven over 51 mph and that diesel fuel costs $1.44 a gallon, find the following.

(a) The steady speed that will minimize the cost of the fuel for a 520 mile trip.

x=

(b) The steady speed that will minimize the total cost of the trip if the driver is paid $14 an hour.
x=

(c) The steady speed that will minimize the total cost of a 680 mile trip if the driver is paid $18 an hour.
x=

Explanation / Answer

(a) Amount of fuel used is proportionate to (676/x)+x, so it is sufficient to find the +ve value of x which minimises (676/x)+x.
Let f(x) = a/x + x
df/dx = 1 - a/x^2
f is minimised when 1 - a/x^2 = 0
a/x^2 = 1
x = sqrt a
If a = 676 then x = 26 --------ANSWER 1

(b) distance = 530
Time = distance/speed
So if the speed = x and the distance = 520 then the time taken = 520/x
and the driver is paid (520/x) * 14 = 7280/x

The cost of the fuel = 0.0054796(676/x+x) gal/mile * 520 miles * $1.44/gal
= 4.103[(676/x)+x]

So the total cost of the trip =
= 7280/x + 4.103[(676/x)+x]  
= 4.103[(1769.43/x)+(676/x)+x]
= 4.103[(2445.43/x)+x]

The total cost of the trip is proportional to (2445.43/x)+x, so it suffices to find the positive value of x which minimises (2445.43/x)+x.

f(x) = a/x + x is minimised when x = sqrt a = sqrt(2445.43) =49.45

When the driver is paid $14 an hour, the total cost of the trip is minimised at approx. 49.45 mph.

(c) This part is exactly same as part (b), with just different numbers given . so plug in them as your practice and get the answer

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