Answer:
The draw in the attached figure
Step-by-step explanation:
Let
L-----> the length of the rectangle
W----> the width of the rectangle
we know that
The perimeter of a rectangle is equal to
[tex]P=2(L+W)[/tex]
we have
[tex]W=23\ ft[/tex]
so
[tex]P=2L+46[/tex] -----> equation A
[tex]P\leq 236\ ft[/tex] ----> inequality B
substitute equation A in the inequality B and solve for L
[tex]2L+46\leq 236[/tex]
[tex]2L\leq 236-46[/tex]
[tex]2L\leq 190[/tex]
[tex]L\leq 95\ ft[/tex]
The maximum possible value of L is 95 ft
therefore
The rectangle could be
Length 95 ft
Width 23 ft
see the attached figure to see the draw