Homework Help Forum: Math

Posted by Salman on Saturday, November 7, 2009 at 3:15pm.

A fence 4 feet tall runs parallel to a tall building at a distance of 2 feet from the building. What is the length of the shortest ladder that will reach from the ground over the fence to the wall of the building?

• Math - Reiny, Saturday, November 7, 2009 at 4:09pm

  Did you make a diagram? Let the distance from the fence to the foot of the ladder be x feet. Let the top of the ladder by y feet above the ground. Let L be the length of the ladder. L^2 = (x+2)^2 + y^2 by similar triangles: 4/x = y/(x+2) so L^2 = (x+2)^2 + 16(x+2)^2/x^2 = (x+2)^2(1 + 16/x^2) 2L(dL/dx) = (x+2)^2(-32/x^3) + (1+16/x^2)(2)(x+2) = (x+2)(-32(x+2)/x^3 + 2(1+16/x^2) For a max/min of L, dL/dx = 0 so (x+2)(-32(x+2)/x^3 + 2(1+16/x^2) = 0 x = -2 clearly cannot be a solution , so (-32(x+2)/x^3 + 2(1+16/x^2) = 0 -32/x^2 - 64/x^3 + 2 + 32/x^2 = 0 2 = 64/x^3 x^3 = 32 x = 32^(1/3) carefully sub x = 32^(1/3) back into the L^2 equation to find L. (I got 8.32 feet)

  
  

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