Calculus
posted by Z32 .
A fence 5 feet tall runs parallel to a tall building at a distance of 7 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? [Hint: Determine the length of a ladder that touches the building, fence, and ground as a function of the acute angle the ladder makes with the ground.]

make a sketch showing the ladder touching the fence and making contact with the wall.
let the angle at the base of the ladder be ß
draw a horizontal from the top of the fence to the wall, then the angle the ladder makes with that line is ß, giving up 2 similar rightangled triangles
the ladder has length L1 above the fence and L2 below the fence
cosß = 7/L1 > L1 = 7secß
sinß = 5/L2 > L2 = 5cscß
so L = 7secß + 5sinß = 7(cosß)^1 + 5(sinß)^1
dL/dß = 7(cosß)^2(sinß)  5(sinß)^2(cosß)
simplifying and setting equal to zero gives me
7sinß/(cosß)^2 = 5cosß/(sinß)^2
crossmultiplying:
7(sinß)^3 = 5(cosß)^3 or
(tanß)^3 = 5/7
tanß = (5/7)^(1/3) = .8939..
ß=41.7936 degrees
sub that back into L1 and L2, add them up , ....