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March 5, 2015

March 5, 2015

Posted by **Anonymous** on Friday, October 28, 2011 at 9:20pm.

- Calculus -
**Reiny**, Friday, October 28, 2011 at 11:30pmMake a diagram

let the foot of the ladder be x ft from the fence

let the ladder reach y ft above the ground

I see similar triangle so set up a ratio

4/x = y/(x+6)

xy = 4x+24

y = (4x+24)/x

let the length of the ladder be L

L^2 = (x+6)^2 + y^2

= (x+6)^2 + [(4x+24)/x]^2

2L dL/dx = 2(x+6) + 2[(4x+24)/x] (x(4) - (4x+24))/x^2

= 2(x+6) - 8(x+6)(-24)/x^3

= 0 for a min of L

2(x+6) - 192(x+6)/x^3 = 0

times x^3

2x^3(x+6) - 192(x+6) = 0

2(x+6)(x^3 - 96) = 0

x = -6 , which makes no sense

or

x = 96^(1/3) , (which is the cuberoot of 96)

x = 4.57886

sub back into L^2 = 197.3

L = 14.05 m

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