A truck has a minimum speed of 15 mph in high gear. When traveling x x mph, the

Question

A truck has a minimum speed of 15 mph in high gear. When traveling x x mph, the truck burns diesel fuel at the rate of 0.0073212(1024x +x)galmile 0.0073212(1024x+x)galmile 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 490 mile trip.
x= x=

(b) The steady speed that will minimize the total cost of the trip if the driver is paid $12 an hour.
x= x=

(c) The steady speed that will minimize the total cost of a 540 mile trip if the driver is paid $20 an hour.
x= x=

Explanation / Answer

solution:

a) here the fuel is proportional to

f(x)=(a/x+x)

f'(x)=1-ax^-2

for f to be minimum

f'=0

a=x^2

if a=1024

the x=32

so cost of fuel will be minimised at x=34

b) now for 490 miles trip

Time taken = distance/speed
suppose speed = y and distance = 490 then the time taken will be 560/x
and the driver is paid amount (490/x) * 12 = 5880/x
now the cost of the fuel is given by  0.0073212(1024/x +x)galmile*$1.44*490 mile
=05.1658 [(1024/x)+x]

so total cost of trip= paid to driver + fuel expenses

= 5.1658 [(1024/x)+x] +5880/x =11169.77/x +5.1658x

now or miniise this

d(total cost)/dt=0

5.1658-11169.77x^-2=0

x=46.5

c) now same problem for 540 mile and 20$ per hour

changing the calculations

the driver is paid amount (540/x) * 20 = 10800/x

the cost of the fuel is given by  0.0073212(1024/x +x)galmile*$1.44*540 mile
=05.6929 [(1024/x)+x]

now totALcost=  = 5.6929 [(1024/x)+x] +10800/x =16629.52/x +5.1658x

for minimise

its derivative=0

5.1658 - 16629.52x^-2=0

x=56.73

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