Answer:
The draw in the attached figure
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
![P=2(L+W)](https://img.qammunity.org/2020/formulas/mathematics/high-school/zp0cq8kh3lxm1bcfeu0a3nf7jca129894r.png)
we have
![W=23\ ft](https://img.qammunity.org/2020/formulas/mathematics/high-school/7gmbyiialr4nx5hp6mbobni5pi1u2yxmqf.png)
so
-----> equation A
----> inequality B
substitute equation A in the inequality B and solve for L
![2L+46\leq 236](https://img.qammunity.org/2020/formulas/mathematics/high-school/tn74w4rlowlt32pmylvu3vr4k4o2d6jay9.png)
![2L\leq 236-46](https://img.qammunity.org/2020/formulas/mathematics/high-school/zrrptt8binnh7eagrgdkqvrq8945d7hhv1.png)
![2L\leq 190](https://img.qammunity.org/2020/formulas/mathematics/high-school/mk6b5r776p2kcp6dky6l13wvsjwl2rcrvy.png)
![L\leq 95\ ft](https://img.qammunity.org/2020/formulas/mathematics/high-school/zv7fhbrc8tgbo03utolqiflabq75q3mp96.png)
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