Date: 11/30/2001 at 11:44:30 From: Michelle Rittenhouse Subject: Ladder Problem Two hallways meet to form a right angle. One is 8 ft. wide and the other is 4 ft. wide. What is the longest ladder that can go around the corner where the hallways meet? The ladder is carried horizontally. Disregard the width of the ladder.
Date: 11/30/2001 at 12:37:41 From: Doctor Rob Subject: Re: Ladder Problem Thanks for writing to Ask Dr. Math, Michelle. I drew this diagram: A x+4 C ---o---------------o :\ | : \ | : \ | 8: \ | : \ | : \ | : x \ | ---o-------o |8+32/x E D|\ | | \ | | \ | 32/x| \ | | \ | | \ | | \| F o.......o B | 4 | Let the length of DE be x. Using similar triangles ADE and DBF, you will find that DF = 8*4/x = 32/x. Then using the Pythagorean theorem on triangle ABC, the length L of AB is given in terms of x by L = sqrt([x+4]^2+[8+32/x]^2), = sqrt([x+4]^2*[1+(8/x)^2]), = sqrt(x^2+64)*(x+4)/x. To find the minimum value of L as the length x changes, which is the maximum length the ladder can have, you can differentiate this with respect to x, set that equal to zero, and solve for x. The equation I got after simplifying was (x^3-256)/(x^2*sqrt[x^2+64]) = 0. Solve this for x, and then you can compute the corresponding value of L. My answer came out between 15 and 20 feet. Feel free to write again if I can help further. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/
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