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Math Help - Differentiation Application Problem

  1. #1
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    Differentiation Application Problem

    I think I have this one right but I'm not sure. Here's what I have

    The problem states: A farmer plans to fence in a rectangular pasture containing three adjacent pens of equal size, lined up end-to-end. The total length of fencing needed to enclose the three pens is 1200 yards. Find the maximum total area that all three pens can enclose.

    So I set up the equation as follows

    A=3xy

    6x+4y=1200

    y=\frac{-6x+1200}{4}

    A=3x(\frac{-6x+1200}{4})=\frac{18x^2+3600x}{4}

    Apply Quotient Rule
    A'=\frac{-144x+14400}{16}=-9x+90

    x=100

    so

    y=150

    A=3(100)(150)

    A=45000

    What did I do wrong?

    Thanks
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  2. #2
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    Quote Originally Posted by beanus View Post
    I think I have this one right but I'm not sure. Here's what I have

    The problem states: A farmer plans to fence in a rectangular pasture containing three adjacent pens of equal size, lined up end-to-end. The total length of fencing needed to enclose the three pens is 1200 yards. Find the maximum total area that all three pens can enclose.

    So I set up the equation as follows

    A=3xy

    6x+4y=1200

    y=\frac{-6x+1200}{4}

    A=3x(\frac{-6x+1200}{4})=\frac{18x^2+3600x}{4}

    Apply Quotient Rule
    A'=\frac{-144x+14400}{16}=-9x+90
    First you apparently lost a minus sign between the second to last and last line. Also, notice that 16 is a constant, so you don't use the quotient rule. So....
    \displaystyle A = \frac{-18x^2 + 3600x}{4}

    \displaystyle A' = \frac{-36x + 3600}{4} = -9x + 900

    Now, you still come out with x = 100, so there is damage, but the end result is the same. You still get y = 150, giving A = 45000. Why do you think something is wrong?

    -Dan
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  3. #3
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    Yeah the missing sign was a typo, sorry. I'm wondering if it is right or not because the online math program I'm using says I'm wrong, and I'm pretty sure I'm right. Thanks Dan!
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