
Optimization Problem
Hi! I need some help with this problem. I need the exact steps involved in getting the derivative of the function. As well as the full answer. I have been trying for a long time and can't get past the derivative. I just can't figure out how to transform it. And please explain how you determine if it is a minimum.
Problem: A new cottage is built across the river and 300 m downstream from the nearest telephone relay station. The river is 120 m wide. In order to wire the cottage for phone service, wire will be laid across the river under water, and along the edge of the river above ground. The cost to lay wire under water is $15 per m and the cost to lay wire above ground is $10 per m. How much wire should be laid under water to minimize the cost? Be sure to check that your answer is indeed a minimum.
Let C equal cost
x=length of wire
Equations:
a= sqrt(x^2 + 120^2)
C(x) = 15x +10(300a)
Solution
C(x)= 15x + 10(300(sqrt(x^2 + 120^2)))
This is where I get lost. I don't know how to get the derivative. Can someone please go through each step. I'm just learning calculus and having a hard time with it. Thanks!

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Re: Optimization Problem

Re: Optimization Problem
Thanks so much for the help. It was a great explanation. The one thing that I do not get though is how you went from the function C(x)=......15(sqrt x^2 +120^2) to it being a fraction in the derivative. I assume when you remove the square you end up with 1/2 and then you inverse it to get (sqrt x^2 +120^2) on the bottom but shouldn't it be (sqrt x^2 +120^2)^1/2? I'm missing how you end up putting (sqrt x^2 +120^2) on the bottom.

Re: Optimization Problem
Remember for f(x) =x^n we have f'(x) = n*x^(n1). In this case we have to find the derivative of ( x^2 + 120^2)^1/2 which wil be 1/2 (x^2+120^2)^(1/2 1) * 2x. And 1/2 1 =  1/2