The figure below represents the ships A and B.
The time required to travel by ship A from A to C is,
t= t= The time required to travel by ship
B is,
t= t= From above two equations,
= xA=2xB The shortest distance is,
AB2=xA2+(200−x2B)2 AB=√(2xA)2+(200−x2B)2 AB=√5xB2−400xB+40000⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(I) For the shortest distance,
(AB)=0 0=(10xB−400)⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(II) xB=40mAgain differentiate equation (II).
(AB)>0atx=40m The distance always greater than zero
Substitute the value in equation (I).
AB=√5×402−400×40+40000 AB=80√5km The time after which the distance between the ships is
shortest is,
t===4hr