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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