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Math Help - Intermediate Algebra

  1. #1
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    Intermediate Algebra

    A tour organizer serves 450 customers in a day, if they charge 10 dollars per person. The company estimates that it will lose 10 customers for every increase of 0.50 dollars in the fare. Find the fare that would yield the maximum income to the company.
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    Re: Intermediate Algebra

    Can you get an equation for the income?
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    Re: Intermediate Algebra

    let the fare be increased by x times half dollar.....
    therefore equation becomes

    <br />
f\left( x \right) = (450 - 10x)(10 + x/2) = 4500 - 10x + 225x - 5{x^2} = 4500 + 215x - 5{x^2}\\<br />
\frac{d}{{dx}}(f\left( x \right)) =  - 10x + 215\\<br />
\frac{d}{{dx}}(f\left( x \right)) = 0\\<br />
 - 10x + 215 = 0\\<br />
x = 21.5\\<br />
f(21.5) = 6811.25\[\max \]<br />

    therefore fare which yields max profit is 10+21.5/2=10+10.75=20.75$
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    Re: Intermediate Algebra

    Just neatening things up a bit:

    A bit of LaTeX coding advice: Unless you are writing a matrix or array avoid using the returns in the tex lines. That will get rid of those pesky < br/ >s. I also got rid of the \\, but I'm not sure it did anything.

    f\left( x \right) = (450 - 10x)(10 + x/2) = 4500 - 10x + 225x - 5{x^2} = 4500 + 215x - 5{x^2}

    \frac{d}{{dx}}(f\left( x \right)) =  - 10x + 215

    \frac{d}{{dx}}(f\left( x \right)) = 0

     - 10x + 215 = 0

    x = 21.5

    f(21.5) = 6811.25\[\max \]

    therefore fare which yields max profit is 10+21.5/2=10+10.75=20.75$

    -Dan
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