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Math Help - Optimization The ladder problem

  1. #1
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    Optimization The ladder problem

    A two-dimensional contractor would like to take a ladder down a hallway, but must negotiate a corner. The dimensions of the hallway are illustrated in Figure below. That is, one width is 8 feet, the other 5 feet. What is the longest ladder that the contractor can successfully get around the corner and through the hallway?



    1. Find a formula for l(t)=x+y:

    2. What is the length of the longest ladder that can fit when the angle is p/6 radians (30 degrees)?

    3. A 20 foot ladder will get stuck on its way around the corner of the hallway - at one angle if carried through the 8' hallway, and at another angle through the 5' hallway. What are these two angles?

    4. If a ladder is longer than this minimum value, there will be angles for which it won't fit around the corner. For ladders shorter than this minimum value this won't be the case. Use this to find the length of the longest ladder the contractor can carry around the corner.


    I am so confused by this problem, I don't even know what the formula is, is it l(t) = 5/cos(t) + 8/sin(t)? can someone help solve this, any help is greatly appreciated
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  2. #2
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by Lunar View Post
    A two-dimensional contractor would like to take a ladder down a hallway, but must negotiate a corner. The dimensions of the hallway are illustrated in Figure below. That is, one width is 8 feet, the other 5 feet. What is the longest ladder that the contractor can successfully get around the corner and through the hallway?



    1. Find a formula for l(t)=x+y:

    2. What is the length of the longest ladder that can fit when the angle is p/6 radians (30 degrees)?

    3. A 20 foot ladder will get stuck on its way around the corner of the hallway - at one angle if carried through the 8' hallway, and at another angle through the 5' hallway. What are these two angles?

    4. If a ladder is longer than this minimum value, there will be angles for which it won't fit around the corner. For ladders shorter than this minimum value this won't be the case. Use this to find the length of the longest ladder the contractor can carry around the corner.


    I am so confused by this problem, I don't even know what the formula is, is it l(t) = 5/cos(t) + 8/sin(t)? can someone help solve this, any help is greatly appreciated
    Let's separate the length of the ladder into L = A + B, where B is the portion to right (looks a little longer in your picture) and A is the portion to the left.

    Both A and B serve as hypotenuses to two right triangles. The verify that the lower left angle of both triangles is t.

    In the lower right triangle: sin t = \frac{8}{B} hence B = \frac{8}{sin t} = 8 csc t.

    Do the same for the upper left triangle:

    cos t = \frac{5}{A} hence A = \frac{5}{cos t} = 5 sec t

    Combine the function that is to be optimized into L = A + B = L(t) = 8 csc t + 5 sec t

    I hope this helps you get started with this problem! Good luck!
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