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1234mutu Group Title

OPTIMIZING--Please help A Norman window has the shape of a semicircle atop a rectangle so that the diameter of the semicircle is equal to the width of the rectangle. What is the area of the largest possible Norman window with a perimeter of p feet?

  • 6 months ago
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  1. theEric Group Title
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    You're not doing derivatives, are you?

    • 6 months ago
  2. 1234mutu Group Title
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    I am @theEric , I am really bad at optimizing

    • 6 months ago
  3. theEric Group Title
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    Do you know what derivatives are? I just want to know what level you're at.

    • 6 months ago
  4. 1234mutu Group Title
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    yes sir i do know the derivatives!:) at least supposed to

    • 6 months ago
  5. theEric Group Title
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    Okay! I'm not sure if that's what we need for this or not. I remember this from high school advanced math, but I need to find an example on the internet...

    • 6 months ago
  6. 1234mutu Group Title
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    https://answers.yahoo.com/question/index?qid=20080928213121AAXVcNK i got this one. thankyou for helping me!

    • 6 months ago
  7. 1234mutu Group Title
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    so how would i take the derivative considering so many variables, that part confuse me!:(

    • 6 months ago
  8. theEric Group Title
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    That post on Yahoo! Answers actually mentions that! Look for all of the relationships between the variables. That way you can substitute! |dw:1398212518200:dw| So, you have the perimeter. That doesn't change. You are concerned with the height; that can change. But it's related to the perimeter. You are concerned with the width of the rectangle, which can change. But it's related to other things as well. And you are concerned with the radius of the semicircle atop the rectangle, which also changes. See, the width is always going to be two times the radius. It will be the same as the diameter. The height is still limited by the known perimeter.

    • 6 months ago
  9. theEric Group Title
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    So, what is the perimeter?

    • 6 months ago
  10. theEric Group Title
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    You can use \(h\) for height, \(w\) for width, \(r\) for radius, \(p\) for perimeter, and \(A\) for area. So many variables!

    • 6 months ago
  11. 1234mutu Group Title
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    haha right!, so i would figure equations for permeter and area. solve to h,w,r then plug them in the area equation and take the derivative?

    • 6 months ago
  12. theEric Group Title
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    Yeah! You have constant \(p\), so you don't have to worry about taking a derivative with respect to it. You want to solve for the area \(A\), so you won't be taking the derivative with respect to it. You'll be taking the derivative of it with respect to something else. Your problem lies within having three variables in your formulas, but you only know how to take the derivative with respect to one. So you see if you can find some variables to be "in terms of" the other. For example. You have \(w\) and \(r\), right? Well, looking at the picture, \(w=2r\). So now you have \(w\) "in terms of" \(r\). And you can put \(2r\) in the place of any \(w\). And so all the \(w\)'s are gone and you're left with 2 of 3 variables: \(r\) and \(h\).

    • 6 months ago
  13. theEric Group Title
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    Does that sound okay with you?

    • 6 months ago
  14. 1234mutu Group Title
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    Yess sir, so i go perimeter equaling P^2/(8+2pi), and i got it right!, Thankyou so much!! big help

    • 6 months ago
  15. theEric Group Title
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    Haha, I didn't do a whole lot! If you got it right, I guess you're good! :)

    • 6 months ago
  16. 1234mutu Group Title
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    haha thankyou !! @theEric

    • 6 months ago
  17. theEric Group Title
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    You're welcome! :) Take care!

    • 6 months ago
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