The cost of fuel per km for a truck travelling at a speed v[km/h] is given by #C(v)=v/100+25/v#. (a) What speed result in the lowest fuel cost per km (b) Assume the driver is paid #$40/h#. What speed would give the lowest cost, including wage for 1000Km?

1 Answer
Feb 10, 2018

(a) #50# km/hr (b) #sqrt6500 ~~80 # km/hr

Explanation:

(a) We seek to find #v# that will minimize #C(v) = v/100+25/v# (with #v > 0#

#C'(v) = 1/100 -25/v^2# and

#1/100 -25/v^2 = 0# at #v=50# km.

Use either the first or second derivative test to verify that #C(50)# is a minimum.

(b)

For a trip of 1,000 km we will have

Fuel cost # = 1000(v/100+25/v)#.

The time required is #1000/v# hr, so the driver cost is

#40 (1000/v)#.

The total cost is

#T(v) = 1000(v/100+25/v) + 40 (1000/v)#

# = 10v +65000/v#

We seek #v# to mimimize #T#.

#T'(v) = 10-65000/v^2 = 0# at

#v^2= 6500#