maximizing perimeter of rectangle under sine

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Nancyspe

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Ok, Here's the problem:

A rectangle with length 2k is inscribed in the region between the x-axis and the graph of y=3sin(2x). The value of k which maximizes the perimeter of the rectangle is?

I assume the peak of the sine wave occurs at 0.785 radians (45 degrees) and the amplitude is 3.

P = (2)(3sin(2x)) + 2(0.785-x)

is the eqn. for the perimeter of the rectangle, but I can't figure out how to get the value of (0.785-x) that maximzes it.

Please help.
 
Let the perimeter be P=2x+2y, where x is the length. Then, we have P=2(2k)+2(3sin(2x))=4k+6sin(2x)

You can differentiate this and get P'=12cos(2x)

12cos(2x)=0.

Can you finish up by solving for x, subbing in and finding k?.
 
If I solve that eqn. for x, I get 2x=90 and 270 degree. Throwing out 270, x=45 degrees. The only problem with that is it gives me a value of 3 for y and 0 for x. The answer is actually 0.43 for k. I think there must be another way to get at it, unless I'm doing something wrong.

The thing you have to remember is, that the rectangle will be centered lengthwise around the point x=45 degrees. Therefore, k=the value of x at 45degrees (0.785 radians) - the value of x at the edge of the rectangle. It would help if I could construct a drawing, but I'm not sure how to do that yet.
 
Here's the graph if it helps visualize.
 

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Nancyspe said:
Ok, Here's the problem:

A rectangle with length 2k is inscribed in the region between the x-axis and the graph of y=3sin(2x). The value of k which maximizes the perimeter of the rectangle is?

Are you sure the problem asked for maximum perimeter?

I assume the peak of the sine wave occurs at 0.785 radians (45 degrees) and the amplitude is 3.

P = (2)(3sin(2x)) + 2(0.785-x)

is the eqn. for the perimeter of the rectangle, but I can't figure out how to get the value of (0.785-x) that maximzes it.

Please help.
Click to expand...
 
I double checked the problem - and yes, it asks for the value that maximizes the perimeter. I know most of the time these problems ask for maximum area, but won't maximum area result in dimensions that also yield maximum perimeter? I'm tearing my hair out over this one. I'm trying to get ready for a big exam and this was an example problem that I couldn't work. Not a critical issue, but I'd like to understand it. Thanks in advance for trying to help.
 
Everytime I do it, I get a value of x=0.61548, which means k=0.1695, which is incorrect. The correct answer is 0.43 according to the key, but for the life of me I cannot see how to arrive at it. Unless the book has a typo - which is entirely possible.

I get the same answer whether I arrive at it by taking the derivative or graphing it in my calculator and tracing to the local Maximum and oh, by the way, I get the same answer when I use the formula for either the area or the perimeter, so I guess it is the same x value regardless of whether you are trying to get area or perimeter.
 
I agree with your solution ...

\(\displaystyle P = 2[3\sin(2x)] + 2 \cdot 2\left(\frac{\pi}{4} - x\right)\)

\(\displaystyle \frac{dP}{dx} = 12\cos(2x) - 4\)

setting \(\displaystyle \frac{dP}{dx} = 0\) ...

\(\displaystyle \cos(2x) = \frac{1}{3}\)

\(\displaystyle x = \frac{1}{2} \arccos\left(\frac{1}{3}\right)\)

\(\displaystyle k = \frac{\pi}{4} - \frac{1}{2} \arccos\left(\frac{1}{3}\right) = 0.169918...\)
 
I'll e-mail the publishers and see if there is an error - thanks for your help. I've found a few other minor errors so I wouldn't be surprised.
 
That's what I kept getting also, but I thought I was wrong because the answer was supposed to be .43

It is probably a text mistake.
 
I don't believe you guys are calculating this properly. First, the maximum amplitude of 3 will occur at x=.25 pi radians since the max sine would normally be at .5 pi radians, and this is sin(2x). So, we assume that the midpoint of the width of the rectangle will occur at x=.25 pi radians, therefore the sum of the 2 widths will be 4*(x-.25 pi radians). The length of a side would be 3 sin(2x), so the sum of the 2 lengths would be 6 sin(2x). Therefore the equation for the perimeter would be 4*(x-.25 pi radians) + 6 sin (2x). We want to get the maximum value for this perimeter, so we differentiate, set it equal to 0 and get:
4 + 12 cos (2x)=0 Solving for x we get x=.955316618. Plugging that value in 3 sin (2x), we get the length to be 2.929427125, and since the length was 2k, that means k=1.41421356, which happens to be the square root of 2.
I checked the result by calculating the perimeter using x=.955316618, and then using .96 and .95. Sure enough, it maxed at .955316618, confirming that value as leading to the maximum perimeter, which was 6.336528068 by the way.
 
I agree with your description, except won't it be (.25pi radians - x) instead of the other way around? The edge of the width will be the midpoint (.25pi) less the value of x (measured from x=0 to x=x). You are taking the value of x (the far edge of the rectangle) and then subtracting .25pi from that.
 
Also, according to the answer key, the answer was k=0.43 not 1.41421. And you were using the side of the rectangle that I designated as the width, not the length. How do you designate which side is the length? I was calling the x-axis side the length. I'm waiting to hear back from the developer of the problem.
 
This is rogerstein responding: I chose to let x be above the midpt of the side (which I call the width, I'll get to that in a moment), and so I subtracted the midpt from it, getting half the width. So the total of the two widths is 4*(x-.25pi). Now, I call it the width because the value is 2* (.955316618-.25 pi) which is .339836909. The other side I call the length because when you plug x=.955316618 into 3 sin(2x) you get 2.828427125, which is considerably longer than .339836909. So 2.828427125 is 2k, and thus k=1.41421356 or the square root of 2.
But there's no need to trust me. One can always very easily check an optimized result in differential calculus by simply using values just above and just below your answer. They should each yield results a tiny bit less extreme than your answer. I did that, and my value for x produced a perimeter a bit longer than the values for x just above it and just below. That's why I'm confident of my answer. If you guys are correct then your perimeter should be greater than mine, which again is 6.336528068. What is the perimeter which you get?
 
I think they are defining k (1/2 the length) to be on the x-axis or what you are calling the width. (I agree, it is the shorter side and therefore the width, but when you initially start the problem, it makes more sense to define the x values as 0.25pi plus or minus k, since we have no idea what the midpoint of the y value would be at that point.) If I assume that fact, then solve for a value of k as you have defined x, I get k=0.955316618 - 0.25pi or 0.169918. Exactly what Galactus and I got. And still not the answer they are giving in the answer key of 0.43.

Which HAS to be correct, since if you plug in the values of x when k=0.169918, you get a y value of 2.82846, whereas if you plug in their value of k and get x values to use in the function, you get y values that are NOT identical.

So, if I assume that k is on the shorter side, work the problem, and at the end, figure out that the length is the other side and then say aha, I will divide the longer side by 2 to get k, I will get 2.82846/2 or 1.41423, the same answer you got and STILL not the one in the answer key.

HAS to be an error. So actually we all agree! :shock:
 
P = 2l + 2w, l = 2k = 3sin(2x), w = Pi/2-2x, P(x) = 6sin(2x)+Pi-4x

P ' (x ) = 12cos(2x)-4, x = .61548, l = 2k = 3sin(2x) = 3sin(2*.61548) = 2.8284, k = sqrt(2)

w = Pi/2-2x = Pi/2 - 2(.61548) = .34, not .43.

Hence Max Perimeter has a length of 2sqrt(2) units and a width of .34 units (about).
 
so you're making the assumption that k is related to the height rather than x ... then what is the purpose of the author using 2k for the "length" as stated in the original problem? why not just say k = rectangle's height?
 
I am assuming that we are looking at the graph of 3sin(2x) from 0 to Pi/2.


Then let any point on the graph equal (x,y) = [x, 3sin(2x)]. now we have the length of the rectangle: 3sin(2x).

Symmetry of the sine wave gives the width as Pi/2-2x. The rest is just algebra and a little calculus.
 
Mr. Flim-Flam,

You originally defined 2k = 3sin(2x). Therefore, 2k = 2.82843 and k = 1.41421.

Once again, the same value that all of us have gotten. The only difference is whether you solve for x on the left side of 0.25pi or on the right side of 0.25pi.
 
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