Optimization Problems

Question

A fence 10 feet tall runs parallel to a tall building at a distance of 2 ft from a building.

We wish to find the length of the shortest ladder that will reach from the ground over the fence to the wall of the building.

[A] First, find a formula for the length of the ladder in terms of θ

[B] Now, find the derivative, L'(θ)L′(θ).

..

[C] Once you find the value of θθ that makes L'(θ)=0L′(θ)=0, substitute that into your original function to find the length of the shortest ladder. 

Answer 1

[A] L(θ) = 10/sinθ + 2/cosθ;

[B] L'(θ) = - 10cosθ/sin²θ + 2sinθ/cos²θ;

[C] L'(θ) = 0; - 10cos³θ + 2sin³θ = 0; tan³θ = 5; θ = tan^-1[(5)^1/3];

L_min = 10/sin{tan^-1[(5)^1/3]} + 2/cos{tan^-1[(5)^1/3]} = 15.546 feet

Answer 2

The minimum length of the ladder is 15.546ft

Answer 3

graph the wall as the y axis, the ground base as the x axis

the point where the ladder touches the fence is (2,10)

the point where the ladder touches the wall is (0,20)

the point where ladder touches ground level is (4,0)

the point where the wall touches the ground is the origin (0,0)

the angle is tan^-1(20/4) = tan^-1(5) = about 78.7 degrees

the ladder length = sqr(20^2 +4^2) = sqr(416) = about 20.4 feet

(2,10) is the midpoint of the ladder, with x and y coordinates that are midpoints of the points of where the ladder touches the wall and the ground

construct a right triangle with the ladder length as the hypotenuse = sqr(20^2+4^2) = sqr416

20 = height, 4 = base

any ladder with length less than 20.4 would not be alble to touch both the ground and the wall if it went over the 10 foot high fence