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Math Help - Minimizing Costs (Word Problem)

  1. #1
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    Minimizing Costs (Word Problem)

    This is a word problem that deals with (minimizing costs):

    A power line is to be constructed from a power station at point A to an island at point C, which is 1 mi directly out in the water from a point B on the shore. Point B is 4 mi downshore from the power station at A. It costs $5000 per mile to lay the power line under water and $3000 per mile to lay the line under ground. At what point S downshore from A should the line come to the shore in order to minimize cost? Note that S could very well be B or A (Hint: The length of CS is sqrt(1 + x^2)

    HELP PLEASE!!!
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    Senior Member TriKri's Avatar
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    Quote Originally Posted by tivo1980 View Post
    This is a word problem that deals with (minimizing costs):

    A power line is to be constructed from a power station at point A to an island at point C, which is 1 mi directly out in the water from a point B on the shore. Point B is 4 mi downshore from the power station at A. It costs $5000 per mile to lay the power line under water and $3000 per mile to lay the line under ground. At what point S downshore from A should the line come to the shore in order to minimize cost? Note that S could very well be B or A (Hint: The length of CS is sqrt(1 + x^2)

    HELP PLEASE!!!
    I guess that ABC forms a right-angle triangle and that S can be chosen anywhere along the line AB.

    Let's say x is the distance in mi upshores from B to C.

    The total cost t\ =\ \sqrt{x^2 +1}\cdot \$ 5000 + (4-x)\cdot \$ 3000.

    t\ =\ (x^2 +1)^{0.5}\cdot \$ 5000 + (4-x)\cdot \$ 3000.

    Minimize t:

    \frac{\partial t}{\partial x}\ =\ 0.5\cdot(x^2+1)^{-0.5}\cdot 2\cdot x\cdot \$ 5000 - \$ 3000\ =\ 0

    \frac{2 x}{2\sqrt{x^2+1}}\cdot \$ 5000\ =\ \$ 3000

    2 x\cdot 5\ =\ 2\sqrt{x^2+1}\cdot 3\ (\Rightarrow x \geq 0)

    50 x^2\ =\ 36(x^2+1)

    14 x^2\ =\ 36

    x^2\ =\ \frac{18}{7}

    x\ =\ 3\sqrt{\frac{2}{7}}

    Answer: 4-3\sqrt{\frac{2}{7}}
    Last edited by TriKri; December 8th 2006 at 04:22 PM.
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  3. #3
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    Thank you Trikri.
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  4. #4
    Senior Member TriKri's Avatar
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    I must be really bored.
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    I have a question Trikri. What software did you use to present the math problem the way you did?
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  6. #6
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    Quote Originally Posted by tivo1980 View Post
    I have a question Trikri. What software did you use to present the math problem the way you did?
    Hello, Tivo,

    have a look here: http://www.mathhelpforum.com/math-he...rial-latex.pdf

    EB
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  7. #7
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    Quote Originally Posted by TriKri View Post
    I guess that ABC forms a right-angle triangle and that S can be chosen anywhere along the line AB.

    Let's say x is the distance in mi upshores from B to C.

    The total cost t\ =\ \sqrt{x^2 +1}\cdot \$ 5000 + (4-x)\cdot \$ 3000.

    t\ =\ (x^2 +1)^{0.5}\cdot \$ 5000 + (4-x)\cdot \$ 3000.

    Minimize t:

    \frac{\partial t}{\partial x}\ =\ 0.5\cdot(x^2+1)^{-0.5}\cdot 2\cdot x\cdot \$ 5000 - \$ 3000\ =\ 0

    \frac{2 x}{2\sqrt{x^2+1}}\cdot \$ 5000\ =\ \$ 3000

    2 x\cdot 5\ =\ 2\sqrt{x^2+1}\cdot 3\ (\Rightarrow x \geq 0)

    50 x^2\ =\ 36(x^2+1)

    14 x^2\ =\ 36

    x^2\ =\ \frac{18}{7}

    x\ =\ 3\sqrt{\frac{2}{7}}

    Answer: 4-3\sqrt{\frac{2}{7}}
    Hello, Trikri,

    I don't agree with your result:

    2 x\cdot 5\ =\ 2\sqrt{x^2+1}\cdot 3\ (\Rightarrow x \geq 0)

    100 x^2\ =\ 36(x^2+1)

    x^2\ =\ \frac{36}{64}

    x\ =\frac{6}{8}. The negative result isn't very realistic with this problem.

    Answer: 4-\frac{3}{4}=3.25\text{ miles} from A.

    EB
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  8. #8
    Senior Member TriKri's Avatar
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    Quote Originally Posted by earboth View Post
    Hello, Trikri,

    I don't agree with your result:

    2 x\cdot 5\ =\ 2\sqrt{x^2+1}\cdot 3\ (\Rightarrow x \geq 0)

    100 x^2\ =\ 36(x^2+1)
    Oh, I was a little tired last night ... I must go to bed earlier!

    x\ =\frac{6}{8}. The negative result isn't very realistic with this problem.
    That's why I showed that x \geq 0
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  9. #9
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    Thank you earboth for the document and the solution.
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