Square in A Parabola

**Rimas** · December 26th 2008, 10:35 AM

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.

**Soroban** · December 26th 2008, 12:06 PM

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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