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 2 years 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?
 2 years 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?

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

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

MarcLeclair
 2 years ago
Best ResponseYou've already chosen the best response.0Well 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)

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

Shadowys
 2 years ago
Best ResponseYou've already chosen the best response.1so, you have two unknowns.

MarcLeclair
 2 years ago
Best ResponseYou've already chosen the best response.0but 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 ?

Shadowys
 2 years ago
Best ResponseYou've already chosen the best response.1yup. (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.

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

Shadowys
 2 years ago
Best ResponseYou've already chosen the best response.1definitely looks like it. good luck. :) okay.

MarcLeclair
 2 years ago
Best ResponseYou've already chosen the best response.0Yikes 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

Shadowys
 2 years ago
Best ResponseYou've already chosen the best response.1im getting \(A=9rr^2\) though....

MarcLeclair
 2 years ago
Best ResponseYou've already chosen the best response.0huh 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

MarcLeclair
 2 years ago
Best ResponseYou've already chosen the best response.0therefore (pi)(r)^ 2 * ( (186(pi)r) / 2 * 2r)

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

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

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

MarcLeclair
 2 years ago
Best ResponseYou've already chosen the best response.0there you go, thats where i screwed up... god I'm retarded...

Shadowys
 2 years ago
Best ResponseYou've already chosen the best response.1lol nah, i think it's because my drawing is damn ugly. lol

MarcLeclair
 2 years ago
Best ResponseYou've already chosen the best response.0ah! got it hahahaha thanks a lot mate! :D
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