1. ## Optimization Problem

A rectangle PQRS is inscribed in the region between the x-axis and the part of the graph y= cos 4x specified by -pi/8 < x< pi/8. ( the < sign is less than or equal to). Determine the coordinates of P for which the perimeter of PQRS is a maximum.

Im not even sure where to begin...

2. Originally Posted by a.a
A rectangle PQRS is inscribed in the region between the x-axis and the part of the graph y= cos 4x specified by -pi/8 < x< pi/8. ( the < sign is less than or equal to). Determine the coordinates of P for which the perimeter of PQRS is a maximum.

Im not even sure where to begin...
Did you draw a diagram?

the height of the rectangle is cos(4x), the length of the rectangle is 2x (we denote the distance from the y-axis to the right bottom corner of the rectangle to be x)

now that you know the height and length of the rectangle, can you continue?

3. Originally Posted by Jhevon
Did you draw a diagram?

the height of the rectangle is cos(4x), the length of the rectangle is 2x (we denote the distance from the y-axis to the right bottom corner of the rectangle to be x)

now that you know the height and length of the rectangle, can you continue?
$\displaystyle P=2x+2\cos(4x)$...assuming you use the same distance x from both sides which it looks like you do...you can do this considering the eveness of cos(x)

4. Originally Posted by Mathstud28
...you can do this considering the eveness of cos(x)
yes, indeed

5. Originally Posted by Mathstud28
$\displaystyle P=2x+2\cos(4x)$...assuming you use the same distance x from both sides which it looks like you do...you can do this considering the eveness of cos(x)

wouldnt P = 2 cos 4x + 4x? because 2 x is the length and P is 2l + 2w??

6. Originally Posted by a.a
wouldnt P = 2 cos 4x + 4x? because 2 x is the length and P is 2l + 2w??
yes

7. ## something is wrong...

ok so P= 2cos 4x + 4x
then P' = -2 sin 4x + 4
so at max. -2 six 4x + 4 = 0
sin 4x = 2.... where do i go from here?

8. Originally Posted by a.a
ok so P= 2cos 4x + 4x
then P' = -2 sin 4x + 4
so at max. -2 six 4x + 4 = 0
sin 4x = 2.... where do i go from here?
Do what you usually do..solve for 0...test that value in second derivative...if you forget this stuff look at this page...it has a huge amount of useful calculus review information

http://www.mathhelpforum.com/math-he...-tutorial.html

9. Originally Posted by a.a
ok so P= 2cos 4x + 4x
then P' = -2 sin 4x + 4
so at max. -2 six 4x + 4 = 0
sin 4x = 2.... where do i go from here?
by the way, your derivative is wrong, you need the chain rule to derive cos(4x)

of course what you have has no solutions (sin(4x) never attains 2, its maximum is 1), you must have realized something was off

10. Originally Posted by a.a
do you mean solve for x... how are we supposed to do that? sin inverse of 2 is undefined
$\displaystyle P=2cos(4x)+4x\Rightarrow{P'=-8sin(4x)+4}$ setting that equal to zero we have $\displaystyle x=\frac{arsin\bigg(\frac{1}{2}\bigg)}{4}+\frac{\pi }{2}n,n\in\mathbb{Z}$...could someone second me on that?

11. -8 sin 4x +4 = 0
sin 4x = 1/2
4x = pi/6
x= pi/24
...........................

12. Originally Posted by a.a
-8 sin 4x +4 = 0
sin 4x = 1/2
4x = pi/6
x= pi/24
...........................
................................Yeah.............. .....................I didnt divide 4/8 wrong .........so the answer is $\displaystyle \frac{\pi}{24}+\frac{\pi}{2}n,n\in\mathbb{Z}$

13. Originally Posted by a.a
-8 sin 4x +4 = 0
sin 4x = 1/2
4x = pi/6
x= pi/24
...........................
ok...and now that you have that evaluate $\displaystyle f''\bigg(\frac{\pi}{24}\bigg)$...if its negative you have a max...if its positive you have a min

14. yip... P'' is -ve hence P is concave down and its a max, so for the given domain the value for x would be pi/24..
thanks for all the help... sorri for all the problems with this one...
i need help with another one im gonna post... if you can please take a look
thanks agian

15. Originally Posted by a.a
yip... P'' is -ve hence P is concave down and its a max, so for the given domain the value for x would be pi/24..
thanks for all the help... sorri for all the problems with this one...
i need help with another one im gonna post... if you can please take a look
thanks agian
Anytime!