ROPE OPTIMIZATION PROBLEM

Question

Two vertical poles PQ and ST are secured by a rope PRS going from the top of the first pole to a point R on the ground between the poles and then to the top of the second pole as in the figure. If R is chosen to minimize the length of the rope, Find the ratio

𝜃1/𝜃2

in terms of a and b where a is the length of PQ and b is the length of ST.

Answer 1

Hopefully you were able to draw yourself a picture of the situation, maybe something like the following:

/ |

/ |

| \ / | b

a | \ / |

| \ / |

R

For this problem, without the loss of generality, we'll let a ≤ b (This works because if a > b, just call a, b and call b, a). Also, let R be the distance from the base of PQ, towards ST.

One last point before we start with the calculus: it seems pretty important that we know the distance separating the poles PQ and ST, but the question asks about the ratio of two angles (I'll assume 𝜃1 is the angle PRQ and 𝜃2 is the angle SRT). By your knowledge of similar triangles, if we scale this drawing, those angles don't change. With that in mind, we'll divide all lengths by the distance between the poles so that the poles are now separated by a distance of 1 unit.

Let's let the poles originally be separated by a distance d. Scaling, we get new values for a and b.

u = a/d

v = b/d

r = R/d

Now we're set up. A great place to start optimization problems is by constructing the function we would like to optimize - in this case, the length of the rope. The length of our rope is just the sum of the hypotenuses of these two triangles we've consequently drawn in our figure. Let's write this function L in terms of r.

L(r) = √[u² + r²] + √[v² + (1-r)²]

This is a function of one variable and we're looking for a local minimum (particularly, a local minimum where r ≥ 0... why?), so we should take the derivative and set it equal to zero to find critical points. Taking this derivative requires a few instances of the Chain Rule, which I'll leave as an exercise. Hopefully, after setting this derivative equal to 0, you eventually get to a point (albeit after a good deal of simplification; in this, some cross multiplication and FOIL-ing can go a long way towards cancelling terms), where you can write a quadratic in terms of r:

(u² - v²) r² - 2u²r + u² = 0

Using the quadratic formula, we arrive at 2 possible solutions (notice you can factor (u² - v²) as a difference of squares if you're struggling to simplify):

r = u / (u+v) and r = u / (u-v)

Recall from the beginning, we let a ≤ b, which means u ≤ v. The second expression therefore doesn't work, as we would get an undefined placement or a negative value for r (meaning r is not in between the poles). Therefore, we are left with the first expression for r.

If we'd like, we can multiply the numerator and denominator by d to change u and v back into a and b.

r = a / (a+b)

Now, we can multiply both sides of the equation by d to change r back into R.

R = d* [a / (a+b)]

Great! Now all that's left is to find expressions for 𝜃1 and 𝜃2 in terms of a and b. We can use tangent in both cases since we know the opposite sides of each triangle are a and b, and the adjacent sides are R and

(d-R), respectively. This gives, in terms of a and b:

tan(𝜃1) = (a+b) / d and tan(𝜃2) = (a+b) / d

This means that 𝜃1 = 𝜃2 and thus 𝜃1 / 𝜃2 = 1.

Answer 2

𝜃1/𝜃2 = 1 , So, θ₁ = θ₂. This is follow if you combain one pole with symmetric image of second pole