# Thread: Finding the maximum

1. ## Finding the maximum

Hello, I am a freshman in college at Western MI, I know how to use derivatives to find the minimum and maximum critical values of a function, but I am having a very hard time solving the following problem:

Borrowed from Calculus by Hughes-Hallett, et. al.
"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 ferry uses $200 worth of fuel per hour when cruising at 10 miles per hour. Apart from fuel, the cost of running this ferry is$1050 per hour. At what speed should the ferry travel to minimize the cost per mile traveled?"

So far I have figured the following:

Cost (relative to speed) = 1/5s^3

and therefore:

Cost (per hour) = 1/5s^3 + 1050

I am having a hard time manipulating this equation to give me cost per mile traveled. Please help me!

2. Originally Posted by w7west
Hello, I am a freshman in college at Western MI, I know how to use derivatives to find the minimum and maximum critical values of a function, but I am having a very hard time solving the following problem:

Borrowed from Calculus by Hughes-Hallett, et. al.
"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 ferry uses $200 worth of fuel per hour when cruising at 10 miles per hour. Apart from fuel, the cost of running this ferry is$1050 per hour. At what speed should the ferry travel to minimize the cost per mile traveled?"

So far I have figured the following:

Cost (relative to speed) = 1/5s^3

and therefore:

Cost (per hour) = 1/5s^3 + 1050

I am having a hard time manipulating this equation to give me cost per mile traveled. Please help me!

Ok I found it!

in case you are interested what I did was:

let f(x) = (1/5)x^3 + 1050

let g(x) = f(x) / x

g'(x) = (2(x^3 - 2625)) / (5x^2)

and I just solved for g'(x) = 0 to find the minimum which turned out to be:

5*21^(1/3)