## MarcLeclair Group Title 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? one year ago one year ago

1. MarcLeclair Group Title

2. MarcLeclair Group Title

|dw:1354534273826:dw| thats what i get as a drawing but i cant seem to get the good equations :/

okay, for this question, you have two equations. one is the perimeter, and the other is the area. can you formulate both equations?

4. MarcLeclair Group Title

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)

well, since the circle is on the rectangle, |dw:1354534781253:dw| so, the length of the rectangle is 2r

so, you have two unknowns.

7. MarcLeclair Group Title

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 ?

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.

9. MarcLeclair Group Title

oh dear its going to look ugly though... I'll try it out and see if i need further help

definitely looks like it. good luck. :) okay.

11. MarcLeclair Group Title

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

im getting $$A=9r-r^2$$ though....

13. MarcLeclair Group Title

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

14. MarcLeclair Group Title

therefore (pi)(r)^ 2 * ( (18-6(pi)r) / 2 * 2r)

$$\pi r + 2r + 2y=18$$ $$y=9-\frac{(\pi+2)r}{2}$$ sub into $$A=\frac{\pi r^2}{2} + ry$$

16. MarcLeclair Group Title

wait you should have 4r because if you divide the rectangles width ( or x ) in 2r , than its perimeter is 4r + 2y

yes, but one side is covered by the semicircle, so one of its side isn't in the perimeter.

|dw:1354537063461:dw|

19. MarcLeclair Group Title

there you go, thats where i screwed up... god I'm retarded...

lol nah, i think it's because my drawing is damn ugly. lol

21. MarcLeclair Group Title

ah! got it hahahaha thanks a lot mate! :D

lol you're welcome :)