Given that the two indentical rectangular pens share a common fencon, then the perimeter to be fenced are three sides (including the common fence) with the same length and two sides of the same length.
Let the length of one of the three sides of the same length be x, then the length of one of the two sides of the same length is [tex] \frac{1}{2} (300-3x) [/tex].
The entire stucture is a stack of two identical rectangles which is also rectangular in shape.
The area of a rectangle is given by length times width. i.e.
[tex]Area= \frac{1}{2} x(300-3x)=150x- \frac{3}{2} x^2[/tex]
For maximum area, the derivative of the area with respect to x equals zero.
[tex] \frac{d}{dx} (Area)=150-3x=0 \\ \\ \Rightarrow3x=150 \\ \\ \Rightarrow x= \frac{150}{3} =50[/tex]
The length of one of the two sides of equal length is [tex]\frac{1}{2} (300-3(50))=\frac{1}{2}(150)=75[/tex]
Therefore, the dimensions of each pen that will make their areas as large as possible is 50 ft by 37.5 ft
