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Math Help - Square in A Parabola

  1. #1
    Member Rimas's Avatar
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    Square in A Parabola

    A rectangle PQRS is inscribed as sketched in the region between the X axis and the part of the graph y= cos(4x) specified by -(8/Pi) less then or equal to x less then or equal to (8/pi). Determine the coordinates of P for which the perimeter of PQRS is a maximum.
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  2. #2
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    Hello, Rimas!

    A rectangle PQRS is inscribed as sketched in the region between the [tex]x[/mathy]-axis
    and the part of the graph y\:=\:\cos(4x), on [-8\pi,\,8\pi]
    Determine the coordinates of P for which the perimeter of PQRS is a maximum.
    Code:
                    |
                    *
                *   |   *
             *- - - | - - -*
           * |      |      | *
          *  |y     |     y|  *
             |      |      |
      ---*---+------+------+---*---
       -8π       x  |  x      8π

    The perimeter of the rectangle is: . P \:=\:4x + 2y .where y \:=\:\cos(4x)

    So we have: . P \:=\:4x + 2\cos(4x)


    Set the derivative equal to zero: . P\:\!' \:=\:4 - 8\sin(4x) \:=\:0 \quad\Rightarrow\quad \sin(4x) \,=\,\frac{1}{2}

    . . Hence: . 4x \:=\:\frac{\pi}{6} \quad\Rightarrow\quad x \:=\:\frac{\pi}{24}

    . . Then: . y \:=\:\cos\left(4\cdot\frac{\pi}{24}\right) \:=\:\cos\left(\frac{\pi}{6}\right) \:=\:\frac{\sqrt{3}}{2}


    Therefore: . P\left(\frac{\pi}{24},\:\frac{\sqrt{3}}{2}\right)

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