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January 31, 2015

January 31, 2015

Posted by **Amber** on Tuesday, April 24, 2012 at 1:16am.

- Calc 1 -
**Steve**, Tuesday, April 24, 2012 at 5:04amif h is the height of the ladder on the wall, and x is the distance of the base of the ladder from the wall, then by similar triangles,

x/h = (x-3)/8

h = 8x/(x-3)

Now, the ladder length y is

y^2 = x^2 + h^2 = x^2 + 64x^2/(x-3)^2

2yy' = 2x + 128x/(x-3)^2 - 128x^2/(x-3)^3

y' = (2x(x-3)^3 + 128x(x-3) - 128x^2)/y(x-3)^2

That's messy, but we want y'=0, and as long as the denominator isn't 0, we just need

2x(x-3)^3 + 128x(x-3) - 128x^2 = 0

x = 3 + 4∛3 = 8.769

h = 8x/(x-3) = 12.161

so, y, the ladder, is 15 ft

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