Wednesday

April 1, 2015

April 1, 2015

Posted by **rolan** on Friday, August 10, 2012 at 5:47am.

- math -
**Reiny**, Friday, August 10, 2012 at 9:21ammake a sketch.

let x be the distance for the wall to where the beam touches the ground

let the height of the beam along the building be y+8 ft

by similar triangles

y/27 = 8/x

xy = 216 -----> y = 216/x

let L be the length of the beam

L^2 = (8+y)^2 + (27+x)^2

= (8 + 216/x)^2 + (27+x)^2

2L dL/dx = 2(8+216/x)(-216/x^2) + 2(27+x)

= 0 for a min of L

2(8+216/x)(-216/x^2) + 2(27+x) = 0

(8+216/x)(-216/x^2) + (27+x) = 0

-1728/x^2 - 46656/x^3 + 27+x = 0

times x^3

-1728x - 46656 + 27x^3 + x^4 = 0

x^3(x+27) - 1728(x+27) = 0

(x^3 - 1728)(x+27) = 0

x = 12 or x = -27, the last one is a "silly" answer

if x=12

L^2 = (8+18)^2 + (27+12)^2

= 2197

L = √2197 = 46.87

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