Finding the Width of a CorridorDate: 11/13/95 at 19:42:41 From: Anonymous Subject: The differential, extremas A steel girder 27 ft long is moved horizontally along a passageway 8 ft wide and into a corridor at right angles to the passageway. How wide must the corridor be for the girder to go around the corner? Neglect the horizontal width of the girder. Date: 11/15/95 at 16:20:36 From: Doctor Ken Subject: Re: The differential, extremas Hello! As your subject line indicates, this is a problem that will use Calculus and local min/max stuff. Here's a hint on how you might set it up: | | | | | / | | / | | / | | / | | / | | / | _____________________|/ | _ / 27 ft. | | / | | / | 8 ft. / | | / | | __________/_________________________| - |---- x ft. ---| So you want to find the minimum value of x such that the girder can still go around the corner, right? Well, for any value of x, there will be a "sticking point" where the girder will fit most tightly, i.e. where the girder's ends are closest to the walls of the corridors as the girder travels around the bend. So we want to find the value of x where this sticking point makes both ends of the ladder exactly touch the walls. See if you can find some similar triangles to help you out. From here I'll direct you to a sketch I made using "The Geometer's Sketchpad." If you have access to a Macintosh, you can look at the sketch by downloading a demo of the program from the URL http://mathforum.org/dynamic.html , and then you can download the sketch I made by using the URL http://mathforum.org/dr.math/sketches/girder.gsp Enjoy! -Doctor Ken, The Geometry Forum |
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