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Maxima and minima word problem

  1. Dec 18, 2012 #1
    1. The problem statement, all variables and given/known data
    A corridor of with a is at right angles to a second corridor of width b.
    A long, thin, heavy rod is to be pushed along the floor from the first corridor into the second.
    What is the length of the longest rod that can get around the corner?


    2. Relevant equations

    I have no idea how to set this equation up :S

    3. The attempt at a solution

    Final answer should be ((a^(2/3)+b^(2/3))^(3/2)

    HELP!! I don't know where to start!
     
  2. jcsd
  3. Dec 18, 2012 #2

    Ray Vickson

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    Start by drawing a diagram of a rod that is going around the corner. What do you see?
     
  4. Dec 18, 2012 #3

    HallsofIvy

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    The longest rod that will go around the corner is the one that is just making contact with the two outside walls when it makes contact with the corner. After you drawn you diagram, you should see several similar right triangles.
     
  5. Dec 18, 2012 #4

    Ray Vickson

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    That won't work. You can take a 1000 mile long rod, and it can be made to touch the outside walls and an inside corner, but it won't go around the corner.

    I think this is actually a somewhat challenging calculus problem, where we require that for any two contact points on the outer walls, the inner corner is a non-negative distance from the rod; so, if the outer walls are the x and y axes, and the ends of the rod are at (u,0) and (0,v) (with u^2 + v^2 = r^2) we require that the point on the rod at (a,y(a)) be at or below (a,b), and that the point (x(b),b) on the rod be to the left of (a,b). Thus, the problem is
    max z, subject to z = u^2 + v^2, y(z) <= b and x(b) <= a. Here, z = r^2, and x(a), y(b) involve u, v.
     
  6. Dec 18, 2012 #5
    We just started doing calculus last week and this was given as a bonus question if anyone can give me a instruction on how to do this question please let me know as I have got no experience in calculus. :(
     
    Last edited: Dec 18, 2012
  7. Dec 18, 2012 #6
    Ok I got down to part where I represented rod length by drawing a big triangle with x and y. So then rod length = SQr of x^2 + y^2 then l prime = 1/2 (x^2+y^2)^(-1/2)* (2x+2yy')
     
  8. Dec 18, 2012 #7
    Substitute y= ax/ x-b
     
  9. Dec 18, 2012 #8

    haruspex

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    If your calculus has reached differentiating trig functions, I would take the angle the rod makes to one of the walls as the independent variable. When the rod touches both walls and the corner, you can write down the length of the rod as a function of that angle and the two corridor widths. Then it's just a matter of finding the angle for which the derivative is 0.
     
  10. Dec 18, 2012 #9

    Ray Vickson

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    Another approach: assume one outer wall is the +x axis and the other outer wall is the +y axis. For a rod of length r, let t0 = angle OAC at which end A is on the +x-axis and the other end B is at the inner corner C = (a,b) (here, O = origin). As we push the end A towards the origin the angle t increases and the other end (B) starts to move up and to the left, getting closer to touching the y-axis. As we change the angle t the other end B traces out a curve in (x,y) space; the curve starts at point C and moves up and to the left initially, then up and to the right again for larget t. We can set up a calculus problem to find the minimum x-value on that curve; that will be the point of closest approach of B to the y-axis. Now find the rod length r that makes the minimum x on the curve equal to zero; that is, which makes the B-curve just graze the y-axis at one point.
     
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