Ladders 1 and 2, denoted L1 and L2, respectively, will rest along two walls (taken to be perpendicular to the ground), and they will intersect at a point O = (a,s), a height s from the ground. Find the largest s such that this is possible. Then find the width of the alley, w = a+b, in terms of L1, L2, and s. This diagram is not to scale.
Without loss of generality, let L2 >= L1.
Observe that triangles AOB and DOC are similar. Let r be the ratio of similitude, so that x=ry. Consider right triangles CAB and ACD. By the Pythagorean theorem, L1^2 - x^2 = L2^2 - y^2. Substituting x=ry, this becomes y^2(1-r^2) = L2^2 - L1^2. Letting L= L2^2 -L1^2 (L>=0), and factoring, this becomes
Now, because parallel lines cut L1 (a transversal) in proportion, r = x/y = (L1-n)/n, and so L1/n = r+1. Now, x/s = L1/n = r+1, so ry = x = s(r+1). Solving for r, one obtains the formula r = s/(y-s). Substitute this into (*) to get
NOTE: Observe that, since L>=0, it must be true that y-2s>=0.
Now, (**) defines a fourth degree polynomial in y. It can be written in the form (by simply expanding (**))
L1 and L2 are given, and so L is a constant. How large can s be? Given L, the value s=k is possible if and only if there exists a real solution, y', to (***), such that 2k <= y' < L2. Now that s has been chosen, L and s are constants, and (***) gives the desired value of y. (Make sure to choose the value satisfying 2s <= y' < L2. If the value of s is "admissible" (i.e., feasible), then there will exist exactly one such solution.) Now, w = sqrt(L2^2 - y^2), so this concludes the solution.
L1 = 11, L2 = 13, s = 4. L = 13^2-11^2 = 48, so (***) becomes