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The crosssection of a tunnel has the form
of a rectangle surmounted by a semicircle.
The perimeter of this cross section is 18 meters.
For what radius of the semicircle will
the crosssection have maximum area?
 one year ago
 one year ago
The crosssection of a tunnel has the form of a rectangle surmounted by a semicircle. The perimeter of this cross section is 18 meters. For what radius of the semicircle will the crosssection have maximum area?
 one year ago
 one year ago

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MarcLeclairBest ResponseYou've already chosen the best response.0
dw:1354534273826:dw thats what i get as a drawing but i cant seem to get the good equations :/
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
okay, for this question, you have two equations. one is the perimeter, and the other is the area. can you formulate both equations?
 one year ago

MarcLeclairBest ResponseYou've already chosen the best response.0
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)
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
well, since the circle is on the rectangle, dw:1354534781253:dw so, the length of the rectangle is 2r
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
so, you have two unknowns.
 one year ago

MarcLeclairBest ResponseYou've already chosen the best response.0
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 ?
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
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.
 one year ago

MarcLeclairBest ResponseYou've already chosen the best response.0
oh dear its going to look ugly though... I'll try it out and see if i need further help
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
definitely looks like it. good luck. :) okay.
 one year ago

MarcLeclairBest ResponseYou've already chosen the best response.0
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
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
im getting \(A=9rr^2\) though....
 one year ago

MarcLeclairBest ResponseYou've already chosen the best response.0
huh how, i get this (186(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
 one year ago

MarcLeclairBest ResponseYou've already chosen the best response.0
therefore (pi)(r)^ 2 * ( (186(pi)r) / 2 * 2r)
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
\(\pi r + 2r + 2y=18\) \(y=9\frac{(\pi+2)r}{2}\) sub into \(A=\frac{\pi r^2}{2} + ry\)
 one year ago

MarcLeclairBest ResponseYou've already chosen the best response.0
wait you should have 4r because if you divide the rectangles width ( or x ) in 2r , than its perimeter is 4r + 2y
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
yes, but one side is covered by the semicircle, so one of its side isn't in the perimeter.
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
dw:1354537063461:dw
 one year ago

MarcLeclairBest ResponseYou've already chosen the best response.0
there you go, thats where i screwed up... god I'm retarded...
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
lol nah, i think it's because my drawing is damn ugly. lol
 one year ago

MarcLeclairBest ResponseYou've already chosen the best response.0
ah! got it hahahaha thanks a lot mate! :D
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
lol you're welcome :)
 one year ago
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