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Math Help - Optimization problem?????

  1. #1
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    Optimization problem?????

    a rectangle is to be inscribed in a right triangle having sides 6 inches, 8 inches, and 10 inches. Determine the dimensions of the rectangle with greatest area.

    I recently tried this question and the answer was found by finding the slope and then using the first and second derivatives of the area but it turns out that I was wrong.
    Now I would like to know how to solve this by using proportions and the derivative tests. So if there are any suggestions whatsoever, thank you because I really don't understand it.
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  2. #2
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    draw a set of coordinate axes ... also, sketch the line y = -\frac{4}{3}x + 8

    note that the line forms a 6-8-10 right triangle in quadrant I.


    inscribe a rectangle in the triangle with sides parallel to the coordinate axes.

    area of the rectangle is ...

    A = xy

    A = x\left(8 - \frac{4}{3}x\right)

    find \frac{dA}{dx} and maximize.
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  3. #3
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    Hello, ciciasi!

    A rectangle is to be inscribed in a right triangle having sides 6, 8, and 10 inches.
    Determine the dimensions of the rectangle with greatest area
    Did you notice that we have a right triangle?
    Code:
          |
     (0,6)*
          |  *
          |     *
          |        *
          + - - - - - *
          |           :  *
          |           :y    *
          |           :        *
      - - + - - - - - + - - - - - * - - -
          |     x               (8,0)

    The equaton of the hypotenuse is: . y \:=\:\text{-}\tfrac{3}{4}x + 6 [1]

    The area of the rectangle is: . A \:=\:xy [2]

    Substitute [1] into [2]: . A \;=\;x\left[\text{-}\tfrac{3}{4}x+6\right]


    And we must maximize the function: . A \;=\;\text{-}\tfrac{3}{4}x^2 + 6x

    Got it?


    Answers: . x = 4,\:y = 3


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    If one side of the rectangle is on the hypotenuse,
    . . we have a different (and much trickier) problem.
    Code:
                  *
                 *   *
                *       *   8
             6 *           *
              * - - - - - - - *
             *|               |  *
            * |               |y    *
           *  |       x       |        *
          *   *   *   *   *   *   *   *   *
          : - - - - - -  10 - - - - - - - :

    Answers: . x = 5,\:y = \tfrac{12}{5}



    Note: In both cases, the maximum area is 12 unitsē.

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  4. #4
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    Unhappy

    Quote Originally Posted by skeeter View Post
    draw a set of coordinate axes ... also, sketch the line y = -\frac{4}{3}x + 8

    note that the line forms a 6-8-10 right triangle in quadrant I.


    inscribe a rectangle in the triangle with sides parallel to the coordinate axes.

    area of the rectangle is ...

    A = xy

    A = x\left(8 - \frac{4}{3}x\right)

    find \frac{dA}{dx} and maximize.
    I already did it this way and thats not how I am supposed to find the answer. its wrong for the way I am supposed to solve it.
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  5. #5
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    ... then how, pray tell, are you supposed to solve it?
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  6. #6
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    Quote Originally Posted by skeeter View Post
    ... then how, pray tell, are you supposed to solve it?
    My professor is saying that I am supposed to use proportions with the x and y sides like subtracting the areas of the smaller triangles that are made when the rectangle is inscribed in the right triangle.
    do you understand?
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  7. #7
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    yes, I do understand.

    use your sketch ... let the large triangle base = 8, height = 6.

    area of rectangle = 24 - area of two smaller similar triangles

    A = 24 - \frac{1}{2}xy - \frac{1}{2}(8-x)(6-y)

    \frac{y}{x} = \frac{6}{8} = \frac{3}{4}

    3x = 4y

    x = \frac{4y}{3}

    A = 24 - \frac{1}{2}\left(\frac{4y}{3}\right)y - \frac{1}{2}\left(8-\frac{4y}{3}\right)(6-y)<br />

    A = 24 - \frac{2y^2}{3} - \left(\frac{2y^2}{3} - 8y + 24\right)

    A = 8y - \frac{4y^2}{3}

    now find \frac{dA}{dy} and maximize ... you should get y = 3 , then x = 4.

    hope your prof is satisfied with his "back door" method.
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  8. #8
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    Exactly. I knew someone would figure it out.
    Thanks for showing the steps. I hope my pro. appreciates this.
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