# Homework Help: Minimum Cost-Word Problem-Please Check?

1. Jun 15, 2013

### Justabeginner

1. The problem statement, all variables and given/known data

A gas pipeline is to be constructed from a storage tank to a house. A road runs straight from the tank, passing at its closest approach 300 feet away from the house at a point 600 feet away from the tank. Pipe laid along the road costs $8 per foot, while pipe laid off the road costs$10 per foot. What is the minimum cost for which the pipeline can be built?

2. Relevant equations

3. The attempt at a solution
C= 8x + 10y

$y= √((600-x)^2 + 300^2))$
$y= √(360000 - 1200x + x^2 + 90000)$
$y= √(450000 - 1200x + x^2)$

$y'= \frac {1}{2} (450000 - 1200x + x^2)^ \frac{-1}{2}$
$8x + 10 (\frac {1}{2} (450000 - 1200x + x^2)^ \frac {-1}{2}) (-1200 + 2x)= 0$
$-8= 5 (450000 - 1200x + x^2)^ \frac {-1}{2} (-1200 + 2x)$
$-8 (450000 - 1200x + x^2) \frac {1}{2}= 5 (-1200 + 2x)$
$-4 (450000 - 1200x + x^2) \frac {1}{2}= 5 (-600 + x)$
$16 (450000 - 1200x + x^2) \frac {1}{2}= 25( x^2 - 1200x + 360000)$
$9x^2 - 9x- 180000= 0$
$x^2 - x - 200000= 0$
Just wondering, is it right so far? And if it is, how can I solve this quadratic without a calculator? Thanks!

EDIT: I solved the problem again as the above method was wrong, and now I have it correct, but I just don't know how to delete posts so sorry for this waste of space on the boards >_<

Last edited: Jun 15, 2013
2. Jun 15, 2013

### verty

You'll have to be much clearer about what you are doing at each step.