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Calculus - optimizing problem

  1. Jul 30, 2013 #1
    1. The problem statement, all variables and given/known data
    A new cottage is built across the river and 300 m downstream from the nearest telephone relay station. The river is 120m 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 cost?


    2. Relevant equations
    [itex] a^2 + b^2 = c^2 [/itex]


    3. The attempt at a solution
    [itex] C = 15\sqrt{x^2+14400} + 10(300-x) [/itex]
    [itex] C' = 15 \cdot (\sqrt{x^2+14400})' - 10 [/itex]
    [itex] = 15 \cdot \frac{1}{2\cdot\sqrt{x^2+14400}} \cdot(x^2+14400)' -10 [/itex]
    [itex] = \frac{15\cdot (x^2)'} {2\cdot\sqrt{x^2+14400}} -10 [/itex]
    [itex] = \frac{15\cdot2x}{2\cdot\sqrt{x^2+14400}} -10 [/itex]
    [itex] = \frac{15x}{\sqrt{x^2+14400}} -10 [/itex]

    I'm just wondering if someone can look this over and let me know where I'm going wrong? I proceeded to solve for f'(x) = 0 and I'm getting 10.7, so I know I'm doing something wrong.

    I apologize for the use of ' for prime (this was how I was taught in the lesson).

    Thank you very much for your time.
     
  2. jcsd
  3. Jul 30, 2013 #2
    Just double checking the numbers (your method looks fine), I get 107.3.

    You might have typed in 1440 instead of 14400 in your calculator, which would explain the difference in exactly one order of magnitude.

    Just note that this value is x, the distance not traveled above ground. The problem asks for the distance underwater so another calculation is needed after this.
     
  4. Jul 30, 2013 #3
    Oh, okay, well then I guess it was actually the next part that I wasn't sure how to do!

    My attempt from there was:
    [itex] 10 = \frac{15x}{\sqrt{x^2+14400}}[/itex]
    [itex] 100 = \frac{15x}{x^2+14400} [/itex]

    I'm going to stop there for now in case I'm already making the mistake?
     
  5. Jul 30, 2013 #4
    or if it's supposed to be
    [itex] 10\cdot\sqrt{x^2+14400} = 15x [/itex]
     
  6. Jul 30, 2013 #5
    You can keep it as a fraction, but when you square each side of the equation make sure to square the numerator i.e. 15x.

    Your second post is another (easier imo) way of looking at the problem. From here you can divide both sides by 10 and then square the whole problem leaving you with:

    14,400+x2= 9/4 x2

    14,400= 5/4 x2

    11,520= x2

    107.3≈ x
     
  7. Jul 30, 2013 #6
    Wow that's embarrassing.. thank you! I guess it's time for a break
     
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