HELP calculus optimization problem: fitting thin rod through corridor

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SUMMARY

The discussion centers on a calculus optimization problem involving the maximum length of a widthless rigid pole that can be maneuvered around a right-angled corner formed by two corridors of widths 'a' and 'b'. The solution parallels the problem of finding the shortest ladder that can reach a wall while avoiding an obstruction, illustrating that both problems involve minimizing or maximizing the same mathematical principles. The key takeaway is that both scenarios require identifying local maxima or minima of a function, which can be confirmed using the first derivative test, with the second derivative method available for further verification.

PREREQUISITES
  • Understanding of calculus concepts, specifically optimization techniques.
  • Familiarity with the first and second derivative tests for identifying maxima and minima.
  • Knowledge of geometric principles related to right-angled triangles and their properties.
  • Experience with problem-solving in real-world applications of calculus.
NEXT STEPS
  • Study the application of the first derivative test in optimization problems.
  • Explore the second derivative test for confirming local maxima and minima.
  • Investigate geometric interpretations of calculus problems, particularly involving right angles.
  • Practice solving similar optimization problems involving physical constraints and geometrical shapes.
USEFUL FOR

Students and professionals in mathematics, particularly those studying calculus and optimization, as well as engineers and architects dealing with spatial constraints in design and construction.

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I am having trouble conceptualizing a calculus optimization problem.
I can find the answer to the problem by using the procedure but i am quite uncertain of how the equations match up with what's actually going on in the situation!

Problem: What is the max length of widthless rigid pole that can be carried around a corner of two corridors of width a and b meeting at a right angle?

The solution is identical to finding the shortest length of a ladder from the ground to a wall if there's a block in front and blocking the wall.
This 2nd Q. makes sense because we are minimizing the length of the ladder and indeed the math spits out a local min value.

Now why is it that the conditions to finding the longest length is identical to find the shortest ladder??
 
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they are solve for either the local maxima, or local minima of a function, both of which correspond to f'(x)=0, so the solution method is the same.

You could use the 2nd derivative method to check whether it is a max or min, though it is often obvious based on the problem framing
 
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It's because the problems are physically very similar. As you turn around the corner, you want to find a ladder which touches both walls only once (if it touches both walls for an extended period of time, then it really passed through the walls at some point and that's not allowed). If you think about the corner you are trying to navigate as the block and the outside walls of the corner as the wall and floor, you're looking for a ladder which is only capable of touching the floor and wall in one position, not many positions. This is going to be the shortest ladder capable of touching the wall and the floor at the same time.

The more pertinent question though, is why do you think the second question makes sense because it's asking for a minimum, and the first question doesn't because it's asking for a maximum?
 

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