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Math Help - maximize area: works one way, not the other

  1. #1
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    maximize area: works one way, not the other

    The question is:

    A right triangle represents a lot (side a=12m side b=4m). Doug wants to put a rectangular fence in that uses side a and b, and one corner touching side c of the triangle. What ar ethe dimensions that maximize the triangle.

    Well, I did this question with calculus using triangle ratios.... Draw a triangle in the rectangle and use the ratio: (4-y)/4 = x/12

    Now, if I solve that ratio for x and use IT in the area equation, (x)(4-y), then I get a differnt number than if I solve for y and use IT.... plus, it isn't a maximum if I use x
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  2. #2
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    Hello, Mike!

    Your wording is off . . . and your ratio is wrong . . .


    A right triangle represents a lot: side a = 12m, side b = 4m.
    Doug wants to put a rectangular fence in that uses side a and b,
    and one corner touching side c of the triangle.
    What are the dimensions that maximize the area of the rectangle?
    Code:
    -     *
    : 4-y |  *
    :     |     *
    :     |        *
    4     *-----------*
    :     |     x     |  *
    :   y |           |y    *
    :     |           |        *
    -     *-----------*-----------*
          :     x     :   12-x    :
          : - - - - - 12- - - - - :
    We have two similar right triangles.

    . . Hence: . \frac{4-y}{x} \:=\:\frac{y}{12-x} \quad\Rightarrow\quad y \:=\:4 - \frac{x}{3}\;\;{\color{blue}[1]}


    The area of the rectangle is: . A \;=\;xy

    Substitute [1]: . A \:=\:x\left(4-\frac{x}{3}\right) \quad\Rightarrow\quad A \:=\:4x-\frac{x^2}{3}

    . . Then: . A' \:=\:4 - \frac{2x}{3} \:=\:0 \quad\Rightarrow\quad\boxed{ x \:=\:6}

    Substitute into [1]: . y \:=\:4-\frac{6}{3} \quad\Rightarrow\quad\boxed{ y \:=\:2}

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  3. #3
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    Okay, now, I did get those answers using a different ratio (on my retries)

    Ratio:

    (4-y)/4 = (x)/(12)

    and from what I gather... I need to do these questions with 2 triangles... one triangle side must be matching one of the dimensions of the length I want, and the other one, bigger, but proportional
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