Posted by **robert** on Thursday, November 17, 2011 at 11:02pm.

a fence 8 ft tall stands on level ground and runs parallel to a tall building. if the fence is 1 ft from the building, find the length of the shortest ladder that will extend from the ground over the fence to the wall of the building. (hint #1: if L represents the length of the ladder, the quantity L is minimized when the quantity L2 is minimized, so you do not have to concern yourself with thte square root-just minimize L2 and that will minimize L. Hint#2: use similar triangles)

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- calculus -
**Reiny**, Thursday, November 17, 2011 at 11:21pm
Make 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

8/x = y/(x+1)

xy = 8x+8

y = (8x+8)/x

let the length of the ladder be L

L^2 = (x+1)^2 + y^2

= (x+1)^2 + [(8x+8)/x]^2

= x^2+2x+1 + 64 + 128/x + 64/x^2

2L dL/dx = 2x + 2 - 128/x^2 - 128/x^3

= 0 for a min of L

2x + 2 - 128/x^2 - 128/x^3 = 0

multiply by x^3

2x^4 + 2x^3 - 128x - 128 = 0

2x^3(x+1) - 128(x+1) = 0

(x+1)(2x^3 - 128) = 0

x = -1 , not likely

or

2x^3=128

x^3=64

x=4

sub into L^2

L^2 = 5^2 + 10^2 = 125

I minimized L^2

- calculus -
**ad**, Saturday, July 14, 2012 at 2:06am
12.727 ft

- calculus -
**zeeshan amir khan**, Sunday, January 12, 2014 at 7:35am
Make 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

8/x = y/(x+1)

xy = 8x+8

y = (8x+8)/x

let the length of the ladder be L

L^2 = (x+1)^2 + y^2

= (x+1)^2 + [(8x+8)/x]^2

= x^2+2x+1 + 64 + 128/x + 64/x^2

2L dL/dx = 2x + 2 - 128/x^2 - 128/x^3

= 0 for a min of L

2x + 2 - 128/x^2 - 128/x^3 = 0

multiply by x^3

2x^4 + 2x^3 - 128x - 128 = 0

2x^3(x+1) - 128(x+1) = 0

(x+1)(2x^3 - 128) = 0

x = -1 , not likely

or

2x^3=128

x^3=64

x=4

sub into L^2

L^2 = 5^2 + 10^2 = 125

I minimized L^2

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