Assume two pens have identical dimensions, common width = W, and
Length = L.
Perimeter
4L+3W=384
=>
L=(384-3W)/4
Area
A(W)=2*W*L
=2W(384-3W)/4
=192W-(3/2)W^2
To get the maximum area, A'(W)=0
192-3W=0 => W=64 => L=(384-3W)/4=96-48=48
Therefore
Length (along common wall) = W = 64 ft
Width (perpendicular to common wall) = L = 48 ft
Area of each pen = WL = 64*48 = 3072 sq. ft.
Assume no constraints in size of pens, then assume one pen has zero length, the perimeter becomes
2L+3W=384 => L=(384-3W)/2
Area,
A(W)=LW=(384-3W)/2*W=192W-(3/2)W^2
To get maximum area,
A'(W)=0 => 192-3W=0 => W=64 => L= 96
Area (of larger pen+zero for the other) = 64*96 = 6144 sq. ft.
This is unlikely to be the expected answer.
