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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?

  • 2 years ago
  • 2 years ago

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  1. MarcLeclair Group Title
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    @Shadowys

    • 2 years ago
  2. MarcLeclair Group Title
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    |dw:1354534273826:dw| thats what i get as a drawing but i cant seem to get the good equations :/

    • 2 years ago
  3. Shadowys Group Title
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    okay, for this question, you have two equations. one is the perimeter, and the other is the area. can you formulate both equations?

    • 2 years ago
  4. MarcLeclair Group Title
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    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)

    • 2 years ago
  5. Shadowys Group Title
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    well, since the circle is on the rectangle, |dw:1354534781253:dw| so, the length of the rectangle is 2r

    • 2 years ago
  6. Shadowys Group Title
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    so, you have two unknowns.

    • 2 years ago
  7. MarcLeclair Group Title
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    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 ?

    • 2 years ago
  8. Shadowys Group Title
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    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.

    • 2 years ago
  9. MarcLeclair Group Title
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    oh dear its going to look ugly though... I'll try it out and see if i need further help

    • 2 years ago
  10. Shadowys Group Title
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    definitely looks like it. good luck. :) okay.

    • 2 years ago
  11. MarcLeclair Group Title
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    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

    • 2 years ago
  12. Shadowys Group Title
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    im getting \(A=9r-r^2\) though....

    • 2 years ago
  13. MarcLeclair Group Title
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    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

    • 2 years ago
  14. MarcLeclair Group Title
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    therefore (pi)(r)^ 2 * ( (18-6(pi)r) / 2 * 2r)

    • 2 years ago
  15. Shadowys Group Title
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    \(\pi r + 2r + 2y=18\) \(y=9-\frac{(\pi+2)r}{2}\) sub into \(A=\frac{\pi r^2}{2} + ry\)

    • 2 years ago
  16. MarcLeclair Group Title
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    wait you should have 4r because if you divide the rectangles width ( or x ) in 2r , than its perimeter is 4r + 2y

    • 2 years ago
  17. Shadowys Group Title
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    yes, but one side is covered by the semicircle, so one of its side isn't in the perimeter.

    • 2 years ago
  18. Shadowys Group Title
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    |dw:1354537063461:dw|

    • 2 years ago
  19. MarcLeclair Group Title
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    there you go, thats where i screwed up... god I'm retarded...

    • 2 years ago
  20. Shadowys Group Title
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    lol nah, i think it's because my drawing is damn ugly. lol

    • 2 years ago
  21. MarcLeclair Group Title
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    ah! got it hahahaha thanks a lot mate! :D

    • 2 years ago
  22. Shadowys Group Title
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    lol you're welcome :)

    • 2 years ago
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