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Math Help - Business application calculus thread 1?

  1. #1
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    Business application calculus thread 1?

    Find the number of units x that produces a maximum revenue R

    1. R= ((1,000,000x))/((.02x^2+1800))

    Now I think I am supposed to differentiate this then set it to zero to find the number but what would I use quotient rule?


    Find the number of units x that produces the minimum average cost per unit c

    C=3000x-x^2 square root(300-x)

    Since cost is C/x

    I did

    ((3000x-x^2 square root(300-x))/((x))

    I am stuck here the radical is kind of confusing me if I took out the x would I just be left with

    x-square root 300 ? and then I have to take the derivative of that and set it to zero?
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  2. #2
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    Re: Business application calculus thread 1?

    It's a good solution to calculate the first derivative and let f'(x)=0 and indeed use the quotientrule for the first one.
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    Re: Business application calculus thread 1?

    Alright

    I did the quotient rule to get

    ((2000x^2+18 eight zero+40,000x^2))/((.02x^2+1800)^2))


    Now I have to solve this to find x. But I am not sure how?

    Can anyone help me simplify this.
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  4. #4
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    Re: Business application calculus thread 1?

    There's a mistake in the numerator, it has to be:
    f'(x)=\frac{20000x^2+1800000000-40000x^2}{(0,02x^2+1800)^2}=\frac{-20000x^2+1800000000}{(0,02x^2+1800)^2}

    Solve f'(x)=0 and keep in my mind that a fraction becomes zero if the numerator becomes zero.
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    Re: Business application calculus thread 1?

    What strategy could I use to solve the fraction?
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    Re: Business application calculus thread 1?

    I gave you a hint, if you want to solve:
    \frac{-20000x^2+1800000000}{(0,02x^2+1800)^2}=0 \Leftrightarrow -20000x^2+1800000000=0

    This is a quadratic equation ...
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  7. #7
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    Re: Business application calculus thread 1?

    As for my second question

    I have ((3000x-x^2 square root(300-x))/((x))

    But I am slightly confused on simplifying it.
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  8. #8
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    Re: Business application calculus thread 1?

    Is the function:
    f(x)=\frac{(3000x-x^2)\cdot(\sqrt{300-x})}{x}
    ?
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    Re: Business application calculus thread 1?

    Yes sir.
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  10. #10
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    Re: Business application calculus thread 1?

    Do you notice that:
    f(x)=\frac{(3000x-x^2)\cdot(\sqrt{300-x})}{x}=\frac{x(3000-x)\cdot(\sqrt{300-x})}{x}
    =(3000-x)\cdot \sqrt{300-x}

    Use the product rule.
    Last edited by Siron; August 6th 2011 at 01:37 PM.
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  11. #11
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    Re: Business application calculus thread 1?

    f(x)=\frac{(3000x-x^2)\sqrt{300-x}}{x}

    f(x)=\frac{x(3000-x)\sqrt{300-x}}{x}

    f(x)= (3000-x)\sqrt{300-x}

    product rule ...
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  12. #12
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    Re: Business application calculus thread 1?

    Applying the product rule I now have

    - square root(300-x)+(3000-x)((-1))/((2 square root (300-x))=0

    The question is how can I solve this
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  13. #13
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    Re: Business application calculus thread 1?

    Your derivative is correct:
    -\sqrt{300-x}+\frac{x-3000}{2\sqrt{300-x}}=0

    Make a common denominator ...
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    Re: Business application calculus thread 1?

    I used 2 square root(300-x) as common denominator to get

    -600+2x+x-30000=0 Did I set this up correctly?
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  15. #15
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    Re: Business application calculus thread 1?

    Quote Originally Posted by homeylova223 View Post
    I used 2 square root(300-x) as common denominator to get

    -600+2x+x-30000=0 Did I set this up correctly?
    Yes, that's correct.
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