Question 179079
Area of Rectangle of Mr dean's yard  =  Length x Width   =  Ld  x  Wd
Area of Rectangle of Mr Tomlin's yard  =  Length x Width   =  Lt  x  Wt

Because both rectangles are similar that means their lenghts and widths would be propotional.
          
                 
Because  Area of Rectangle of Mr Tomlin's yard = 4 x Area of Rectangle of Mr dean's yard

           Lt  x  Wt   =  4 (Ld  x  Wd)
Then
           Lt  x  Wt   =  2Ld  x  2Wd

This means  Length of Mr Tomlin's yard 2 times of Length of Mr Dean's yard and
            width of Mr Tomlin's yard 2 times of width of Mr Dean's yard. 

the length of Mr dean's yard is 15m   
Then the length of Mr Tomlin's yard is 15x2 = 30 m
the area of Mr Tomlin's yard is 720m square = Lt x Wt   = 30 x Wt
Wt = 720/30   =  24 m   (the width of Mr Tomlin's yard)

Lt = 30 , Wt = 24 
the lenght of fencing of Mr Tomlin's would be 2*(30+24)
and the cost of fencing of Mr Tomlin's would be 2*(30+24)*15 = $1620 

For Mr Dean's yard
Ld = 15 m
Wd = half of the width of Mr Tomlin's =  (1/2)*24 =  12  m
the lenght of fencing of Mr Dean's would be 2*(15+12)
and the cost of fencing of Mr Dean's would be 2*(15+12)*15 = $810