Q.153. Maxima/Minima

Question 153.
Two vertical poles are 10 meters apart standing on flat ground. The pole are 12 meters high, and 20 meters high. Guy wires are attached to the top of each pole and are also attached to a single stake in the ground directly between the poles. Where should the stake be placed so that the combined length of the guy wires will be a minimum? (distance to stake from pole is x).

Answer 153.
Location of the stake is 3.75 m from 12 m pole or 6.25 m from 20 m pole solved as under.

The best way to solve this problem is using the principle of physics according to which light travels from one point to another in minimum time even while travelling through different media. If the medium is the same, then minimum time also means the shortest path. Now let us see how the above problem can be solved easily using this principle.

Let AB be the 12 m pole with A at top and CD be the 20 m pole with C at top. Let P be the stake on the ground on the line joining BD. Now if a ray of light starts from A and after getting reflected at P passes through C, the length APC will be the shortest. We also know that the angle of incidence = angle of reflection. This means that the ΔABP and CDP are similar.
 Let BP = x. Then, DP = 10 - x.
From similarity of ΔABP and CDP ,
(10 - x) / 20 = x / 12 = 10 / 32
=> x = 12 * (10 / 32) = 15/4 = 3.75 m
and 10 - x = 25 / 4 = 6.25 m

Thus, location of the stake is 3.75 m from 12 m pole or 6.25 m from 20 m pole.

Now, let us see how to do it using calculus,
Total length of guy wires
y = AP + PC = √(x^2 + 144) + √[(10 - x)^2 + 400]
=> y = √(x^2 + 144) + √(x^2 - 20x + 500)
=> dy/dx = x / √(x^2 + 144) + (x - 10) / √(x^2 - 20x + 500) = 0
=> x^2 (x^2 - 20x + 500) = (x - 10)^2 * (x^2 + 144)
=> x^4 - 20x^3 + 500x^2 = x^4 - 20x^3 + 244 x^2 - 2880x + 14400
=> 256 x^2 + 2880 x - 14400 = 0
=> 4x^2 + 45 x - 225 = 0
=> x = (1/8) [ - 45 + √(2025 + 3600) ]
=> x = (1/8) (30) = 15/4 = 3.75 m.

Link to YA!