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MarcLeclair

  • 3 years ago

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?

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  1. MarcLeclair
    • 3 years ago
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    @Shadowys

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

  3. Shadowys
    • 3 years ago
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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?

  4. MarcLeclair
    • 3 years ago
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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)

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

  6. Shadowys
    • 3 years ago
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    so, you have two unknowns.

  7. MarcLeclair
    • 3 years ago
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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 ?

  8. Shadowys
    • 3 years ago
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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.

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

  10. Shadowys
    • 3 years ago
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    definitely looks like it. good luck. :) okay.

  11. MarcLeclair
    • 3 years ago
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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

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

  13. MarcLeclair
    • 3 years ago
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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

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

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

  16. MarcLeclair
    • 3 years ago
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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

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

  18. Shadowys
    • 3 years ago
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    |dw:1354537063461:dw|

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

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

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

  22. Shadowys
    • 3 years ago
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    lol you're welcome :)

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