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A fence 4 feet tall runs parallel to a tall building at a distance of 5 feet from the building. What is the length of the short...Asked by REALLY NEED HELP!!!!
A fence 6 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?
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Answered by
Reiny
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
6/x = y/(x+2)
xy = 6x+12
y = (6x+12)/x
let the length of the ladder be L
L^2 = (x+2)^2 + y^2
= (x+2)^2 + [(6x+12)/x]^2
2L dL/dx = 2(x+2) + 2[(6x+12)/x] (x(6) - (6x+12))/x^2
= 0 for a min of L
x+2 + (6x+12)(-12)/x^3 = 0
.. ..
x^4 + 2x^3 - 72x - 144 = 0
x^3(x+2) - 72(x+2) = 0
(x^3 - 72)(x+2) = 0
x = -2 , not possible
or
x = 72^(1/3) , (which is the cuberoot of 72)
sub back into L^2 = ....
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
6/x = y/(x+2)
xy = 6x+12
y = (6x+12)/x
let the length of the ladder be L
L^2 = (x+2)^2 + y^2
= (x+2)^2 + [(6x+12)/x]^2
2L dL/dx = 2(x+2) + 2[(6x+12)/x] (x(6) - (6x+12))/x^2
= 0 for a min of L
x+2 + (6x+12)(-12)/x^3 = 0
.. ..
x^4 + 2x^3 - 72x - 144 = 0
x^3(x+2) - 72(x+2) = 0
(x^3 - 72)(x+2) = 0
x = -2 , not possible
or
x = 72^(1/3) , (which is the cuberoot of 72)
sub back into L^2 = ....
Answered by
REALLY NEED HELP!!!!
THANKS YOU Reiny!!! You are a great help. Turns out my "y" was wrong but I did get the "x"
Answered by
Reiny
Did you get a shortest ladder of 9.81 ft ?
Answered by
REALLY NEED HELP!!!!
I actually got 10.811 ft
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