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Max Min application

  1. Nov 27, 2005 #1
    Heres an intersting problem i would like to see max-min based solutions for. Fine the maximum length of a bar to be transported horizontally around a 90 degree corner. That is, from a corridor of width a to a corridor of width b.
    -----------
    --------l l
    l l

    thanks
     
  2. jcsd
  3. Nov 27, 2005 #2

    HallsofIvy

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    That's a fairly standard calculus problem. Draw the bar just fitting around the corner: a straight line in your picture just touching the inside corner and the two outside walls. Get a fomula for the length of bar as a function the angle it makes with one of the walls. find the angle that minimizes that length. The length at that angle is the maximum length that will fit around the corner.
     
  4. Nov 27, 2005 #3

    mathman

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    Let L=a/sinx + b/cosx. Find x for minimum L. x is the angle between the bar (assuming 0 thickness) and the wall on the a side. L is then the length of the bar that will fit.
     
  5. Nov 27, 2005 #4
    forgive me, but i dont understand why to let L=a/sinx + b/cosx. shouldnt it be 2L=a/sinx + b/cosx? in either case, i get a min of x= arctan of cube root of (a/b). is this normal? my intuition says that the answer should be sqrt(a^2 + b^2). does this angle provide this answer? if so, how? i still believe there is something i have not caught on to. thanks for your patience.
     
  6. Nov 28, 2005 #5

    HallsofIvy

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    No, why should it be? Are you thinking that a/sinx+ b/cosx is the length of the bar that is in each corridor and should be doubled because there are two corridors? No, a/sin x is the length of the bar in the corridor of width a and b/sin x is the length of the bar in the corridor of width b. The length of the entire bar is their sum: L= a/sin x+ b/cos x.

    Yes, [itex]arctan x= ^3\sqrt{\frac{a}{b}}[/itex]

    Now, draw a triangle with legs of length a1/3 and b1/3 (so that the tangent of the angle has is (a/b)1/3). Then the hypotenuse has length [itex]\sqrt{a^{2/3}+ b^{2/3}}[/itex] and sin x= [itex]\frac{a^{1/3}}{\sqrt{a^{2/3}+ b^{2/3}}}[/itex] and cos x= [itex]\frac{b^{1/3}}{\sqrt{a^{2/3}+ b^{2/3}}}[/itex].

    Then the length of the bar is
    [tex]a^{2/3}\sqrt{a^{2/3}+ b^{2/3}}= b^{2/3}\sqrt{a^{2}{3}+ b^{2}{3}}= (a^{2/3}+ b^{2}{3})\sqrt{a^{2/3}+ b^{2/3}}= \left(a^{2/3}+ b^{2/3}\right)^{3/2}[/tex]
     
  7. Nov 28, 2005 #6
    Oh, i did find that the length can be expressed as L=a/sinx or L=b/cosx, but i did not see it as two seperate parts. so i just added the equations together. im afraid i dont see what you mean by "length of the bar in the corridor of width a". shouldnt the length stay the same in both corridors? i am still confused. perhaps we are thinking of seperate diagrams? thanks
     
  8. Nov 28, 2005 #7
    Eureka! i see what you mean now. indeed i was looking at the wrong diagram. i understand what you mean, thats excellent, thank you. however, one comment i might add, is that couldnt we create a triangle with the same tangent value with one side as a, and the other as (ab^2)^1/3 ? doesnt one side have to be the width of the corridor that it is in? (likewise for corridor b). tell me what you think, thanks
     
  9. Nov 28, 2005 #8

    HallsofIvy

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    The two corridors are of different widths. If you take x to be the angle the bar makes with the wall in hall of width a, you have a triangle with one side of length a and hypotenuse of length L1, the length of the portion of the bar that is in that hall: sin(x)= a/L1 so L1= a/sin(x). The angle the wall makes with the wall in the other corridor is the complement of x so sin of that angle is cos(x). Letting L2 be the length of the portion of the bar in that angle we have, by the same argument L2= b/cos(x). The full length of the bar is a/sin(x)+ b/cos(x)

    It's not clear to me where you got (ab^2)^(1/3).
     
  10. Nov 28, 2005 #9
    well, we know that tanx= cube root (a/b). however, the problem implies that one side of the triangle must be a (the width of the corridor). thus, the only way to get this tangent value is if the other adjacent side of the right triangle is what i gave you in the last post (work it out, it is the same as cube root of a/b). my conflict was with your claim that the sides of the triangle must be a^1/3 and b^1/3. it could be that i just did not understand what you meant. but i think that in eah triangle, one side must be a or b no? (depending on which corridor you are in) thanks
     
  11. Jul 15, 2007 #10
    This has been bugging me

    Can somebody help me. This problem (max of L=a/sinx + b/cos x) has been bugging me for about a year. I knew how to do it 25 years ago and it seemed to involve using tanx/2 substitutions and the tan half angle formula for cot. Why is the max of a/sin x+ b/cos x when cot x = cube root of a/b?

    Thanks in Advance
     
  12. Jul 15, 2007 #11

    mathman

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    Take the derivative of L with respect to x and set it equal to 0. You then get:

    acosx/sin2x=bsinx/cos2x

    which leads to tanx=(a/b)1/3
     
  13. Jul 15, 2007 #12

    bel

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    I think a certain issue of the journal of the American Mathematical Society in 1993 discusses a variation of this problem where the bar's or rod's width or diameter is taken into account, sorry I couldn't give a more precise issue identification, but you might want to look into that.
     
  14. Jul 16, 2007 #13

    HallsofIvy

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    Write f(x)= a (sin x)-1+ b(cos x)-1

    Then f'(x)= -a (sin x)-2cos(x)+ b (cos x)-2(sin x)
    Setting that equal to 0 we get
    [tex]\frac{a cos x}{sin^2 x}= \frac{b sin x}{cos^2 x}[/tex]
    dividing both sides by a sin x and multiplying both sides by cos2 x gives
    [tex]\frac{cos^3 x}{sin^3 x}= cot^3 x= \frac{b}{a}[/tex]
    then
    [tex] cot x= ^3\sqrt{\frac{b}{a}}[/tex]
     
  15. Jul 16, 2007 #14
    Thanks for your postings. I'm amazed at how much of pickle my teenage children's homework has gotton me into. Still, it has revitalized my interest in mathematics.
     
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