Question
An 8 foot tall fence separates Larry's yard from Evan's yard.The fence is 3 feet from Larry's house and runs parallel to the side of his house. Larry wants to paint his house and needs to position a ladder extending from Evan's yard over the fence to his house. Assuming the vertical wall of Larry's house is less than 30 feet tall and Evan's yard is at least 20 feet long what is the length of the shortest ladder whose base is in Evan's yard, clears the fence and reaches Larry's house?
Answers
if 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
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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