Question

Wishing to erect 2 vertical poles, 10 m and 15 m on ground with bases 20 m apart fronting his home. The 2 poles are anchored by a cable from each top to a point on the ground between them. Where should he anchor to minimize the length of the cable?

Answer 1

Let `AB=10m and CD=15m` be the heights of two vertical poles situated at `20m` apart on the ground.. Let `O` be the position on the ground at `x` m apart from `B,` where the two cables from the top of two poles are to be anchored so that the total length of the rope `AO+OC` is minimum.

Let `angle OAB=alphaand angle OCD=beta`

Now `sinalpha=(OB)/(AO)=x/sqrt(10^2-x^2)`
and
`sinbeta=(OD)/(OC)=(20-x)/sqrt(15^2+(20-x)^2)`

Now total length of the two cables

`y=OA+OC=sqrt(10^2-x^2)+sqrt(15^2+(20-x)^2)`

`y=sqrt(10^2+x^2)+sqrt(15^2+(20-x)^2)`

`=>y=sqrt(100+x^2)+sqrt(15^2+(20-x)^2)`

Differentiating w r to `x` we get

`(dy)/(dx)=1/2xx(2x)/sqrt(10^2+x^2)-1/2xx(40-2x)/sqrt(15^2+(20-x)^2)`

`=>(dy)/(dx)=x/sqrt(10^2+x^2)-(20-x)/sqrt(15^2+(20-x)^2`

Imposing the condition of minimization of `y` i.e. `(dy)/(dx)=0`

`=>0=x/sqrt(10^2+x^2)-(20-x)/sqrt(15^2+(20-x)^2`

`=>x/sqrt(10^2+x^2)=(20-x)/sqrt(15^2+(20-x)^2`

`=>sinalpha=sinbeta`

`=>alpha=beta`

`=>cotalpha=cotbeta`

`=>10/x=15/(20-x)`

`=>15x=200-10x`

`=>25x=200`

`=>x=8` m .

So the two cables are to be anchored at a distance `8m` from the pole of height `10m`.