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1234mutu
 one year ago
OPTIMIZINGPlease 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?
1234mutu
 one year ago
OPTIMIZINGPlease 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?

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theEric
 one year ago
Best ResponseYou've already chosen the best response.1You're not doing derivatives, are you?

1234mutu
 one year ago
Best ResponseYou've already chosen the best response.0I am @theEric , I am really bad at optimizing

theEric
 one year ago
Best ResponseYou've already chosen the best response.1Do you know what derivatives are? I just want to know what level you're at.

1234mutu
 one year ago
Best ResponseYou've already chosen the best response.0yes sir i do know the derivatives!:) at least supposed to

theEric
 one year ago
Best ResponseYou've already chosen the best response.1Okay! 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...

1234mutu
 one year ago
Best ResponseYou've already chosen the best response.0https://answers.yahoo.com/question/index?qid=20080928213121AAXVcNK i got this one. thankyou for helping me!

1234mutu
 one year ago
Best ResponseYou've already chosen the best response.0so how would i take the derivative considering so many variables, that part confuse me!:(

theEric
 one year ago
Best ResponseYou've already chosen the best response.1That 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.

theEric
 one year ago
Best ResponseYou've already chosen the best response.1So, what is the perimeter?

theEric
 one year ago
Best ResponseYou've already chosen the best response.1You can use \(h\) for height, \(w\) for width, \(r\) for radius, \(p\) for perimeter, and \(A\) for area. So many variables!

1234mutu
 one year ago
Best ResponseYou've already chosen the best response.0haha 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?

theEric
 one year ago
Best ResponseYou've already chosen the best response.1Yeah! 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 \(h\).

theEric
 one year ago
Best ResponseYou've already chosen the best response.1Does that sound okay with you?

1234mutu
 one year ago
Best ResponseYou've already chosen the best response.0Yess sir, so i go perimeter equaling P^2/(8+2pi), and i got it right!, Thankyou so much!! big help

theEric
 one year ago
Best ResponseYou've already chosen the best response.1Haha, I didn't do a whole lot! If you got it right, I guess you're good! :)

1234mutu
 one year ago
Best ResponseYou've already chosen the best response.0haha thankyou !! @theEric

theEric
 one year ago
Best ResponseYou've already chosen the best response.1You're welcome! :) Take care!
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