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#.