Upper a 4-ft-tall fence runs parallel to a wall of a house at a distance of 24 ft. (a) find the length of the shortest ladder that extends from the ground to the house without touching the fence. assume the vertical wall of the house and the horizontal ground have infinite extent. (a) let l be the length of the ladder, x be the distance from the base of the ladder to the fence, d be the distance from the fence to the house and b be the distance from the ground to the point the ladder touches the house. what is the objective function, in terms of x?

(a) shortest ladder length ≈ 35.7 ft (rounded to tenth)

(b) L = (d/x +1)√(16+x²) . . . . where 16 is fence height squared

Step-by-step explanation:

It works well to solve the second part of the problem first, then put in the specific numbers.

We have not been asked anything about "b", so we can basically ignore it. Using the Pythagorean theorem, we find the length GH in the attached drawing to be ...

  GH = √(4²+x²) = √(16+x²)

Then using similar triangles, we can find the ladder length L to be that which satisfies ...

  L/(d+x) = GH/x

  L = (d +x)/x·√(16 +x²)

The derivative with respect to x, L', is ...

  L' = (d+x)/√(16+x²) +√(16+x²)/x - (d+x)√(16+x²)/x²

Simplifying gives ...

  L' = (x³ -16d)/(x²√(16+x²))

Our objective is to minimize L by making L' zero. (Of course, only the numerator needs to be considered.)

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(a) For d=24, we want ...

  0 = x³ -24·16

  x = 4·cuberoot(6) ≈ 7.268 . . . . . feet

Then L is

  L = (24 +7.268)/7.268·√(16 +7.268²) ≈ 35.691 . . . feet

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(b) The objective function is the length of the ladder, L. We want to minimize it.

  L = (d/x +1)√(16+x²)

i belive those are the ones that are under 90 degreese.

what is the answer?