Let #AB=DC=x# meter.Then, #AD=BC=525/x# meter,

and the cost is

#f(x)=11x + 3(x+2*525/x)#

#=14x+3150/x# dollars. Needless to say, the domain is #x>0#.

Differenciate #f(x)#. Note that #d/dx(1/x)=d/dx(x^-1)=-x^-2=-1/x^2#.

#f'(x)=14-3150/x^2#

Now solve #f'(x)=0# to find the extrema.

#14-3150/x^2=0#

#14x^2-3150=0#

#x^2-225=0#

#(x+15)(x-15)=0#

#x=-15,15#

#x# is the length of #AB# and must be positive. So #x=15#.

And, when #0< x< 15#, #f(x)# is decreasing because #f'(x) <0#. When #15< x#, #f(x)# is increasing (#f'(x)> 0#).

Therefore, #f(15)# is the local (and global) minima and its value is #f(15)=14*15+3150/15=420#.

The minimum cost is #420# dollars.

graph{14x+3150/x [-3.15, 30, 400,500]}