Calculus optimization tricky problem?

Question

The cost of fuel to propel a boat through the water(in dollars per hour) is proportional to the cube of the speed. A certain ferryboat uses $100 worth of fuel per hour when cruising at 10 miles per hour. Apart from fuel, the cost of running this ferry is $675 per hour. At what speed should it travel so as to minimize the cost per mile traveled?

Answer 1

x = speed
F(x) = fuel as a function of speed
F(x) = kx^3

then use F = 100 and x = 10 to find k
100 = 1000k
k = 1/10
F(x) = x^3/10

total cost is then:
c(x) = x^3/10 + 675

Now you need cost per mile. Since the function is cost per 1 hour, the speed will equal the distance traveled (ex. 3 mph = 3 miles in 1 hour). Divide the whole cost equation by x to get the average cost per mile, then take the derivative (and set equal to 0) to find the minimum.

C(x) = x^2/10 + 675/x
C'(x) = x/5 - 675/x^2
0 = x/5 - 675x^2
675/x^2 = x/5
3375 = x^3
x = 15

To prove this is a minimum, the second derivative must be positive at this point:
C''(x) = 1/5 + 1350/x^3
C''(15) = 3/5 (confirms it is a minimum)

So, the minimum cost is when the boat travels at 15 mph, giving an average cost of $67.50 per hour.

Answer 2

Fuel Cost =C = ks^3, where k is a constant and s is speed.
100 = k*10^3 so k = .1
So C = .1s^3
Total cost =T = .1s^3+675
dT/ds = .3s^2
So s = 0 miles per hour = minium cost to run boat