anonymous
  • anonymous
The cross-section of a tunnel has the form of a rectangle surmounted by a semi-circle. The perimeter of this cross section is 18 meters. For what radius of the semi-circle will the cross-section have maximum area?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@Shadowys
anonymous
  • anonymous
|dw:1354534273826:dw| thats what i get as a drawing but i cant seem to get the good equations :/
anonymous
  • anonymous
okay, for this question, you have two equations. one is the perimeter, and the other is the area. can you formulate both equations?

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anonymous
  • anonymous
Well that's where I am "chocking". For the Perimeter do i do: 2(Pi)(r) / 2 + (L+2w) Or do I assume the triangle has 4 side, or was my formulation alright? For the area, hum I would presume (Pi)(r^ 2)/2 + (w)(l)
anonymous
  • anonymous
well, since the circle is on the rectangle, |dw:1354534781253:dw| so, the length of the rectangle is 2r
anonymous
  • anonymous
so, you have two unknowns.
anonymous
  • anonymous
but wait ,is that how its drawn actually ? It was a guess of mine. Can't I just divided it in 2 ( so perimeter of rectangle + perimeter of circle)? But with my way i end up with 3 unknown... So I have to change my length = 2r ?
anonymous
  • anonymous
yup. (the word "surmounted" tells me that) yup, try to formulate so you end up with two unknowns only, before subbing one of them into the second eq.
anonymous
  • anonymous
oh dear its going to look ugly though... I'll try it out and see if i need further help
anonymous
  • anonymous
definitely looks like it. good luck. :) okay.
anonymous
  • anonymous
Yikes I'm getting this as the area ( (12)(pi)r^ 3 - 252(pi)r^ 2 +648r ) /2 Derivative gives me some crazy number ( a parabola). When I try to factor i get in the 4 digit number. I find it a bit high x.x
anonymous
  • anonymous
im getting \(A=9r-r^2\) though....
anonymous
  • anonymous
huh how, i get this (18-6(pi)r) / 2 = l And your are is A=(pi)(r)^ 2 * l*w and I reaplce the l with the above expression and w with 2 r
anonymous
  • anonymous
therefore (pi)(r)^ 2 * ( (18-6(pi)r) / 2 * 2r)
anonymous
  • anonymous
\(\pi r + 2r + 2y=18\) \(y=9-\frac{(\pi+2)r}{2}\) sub into \(A=\frac{\pi r^2}{2} + ry\)
anonymous
  • anonymous
wait you should have 4r because if you divide the rectangles width ( or x ) in 2r , than its perimeter is 4r + 2y
anonymous
  • anonymous
yes, but one side is covered by the semicircle, so one of its side isn't in the perimeter.
anonymous
  • anonymous
|dw:1354537063461:dw|
anonymous
  • anonymous
there you go, thats where i screwed up... god I'm retarded...
anonymous
  • anonymous
lol nah, i think it's because my drawing is damn ugly. lol
anonymous
  • anonymous
ah! got it hahahaha thanks a lot mate! :D
anonymous
  • anonymous
lol you're welcome :)

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