Let the initial man be X and second man be Y
Let coordinates of X at t(any time)hrs be (5t,0)
Therefore coordinates of Y at t hrs is [0,{5*(1.5-t)}]
Therefore the distance b/w the two men will be:

[{5*(1.5-t)}^2 + (5t)^2]

 

=[(1.5)^2+50(t^2)-75t]--------------------------(1)

 

now to get the minimum value of t we must use maxima and minima:

 

Differentiating (1)

 

1/4 ( 400t - 300 )

______________                 =   0

( 225+200(t^2)-300t)

 

or 400 t = 300

 or t=3/4 hrs = 3/4 * 60 mins = 45 mins

 

Therefore the time when the two men will be at min distance apart will be:

 

12:45 p.m