Answer
Possible sides (in feet) are in $(0, 6]$ or $[19, 25)$
Work Step by Step
Let $P$ be the perimeter of a rectangle.
As we are given $P = 50$ feet.
The perimeter $P$ of a rectangle is given as follows:
$P = 2L + 2W$
This implies
$50 = 2L + 2W$
$25 = L + W$
$L = 25 – W$ … (i)
Let $A$ be the area of a rectangle.
As we are given $A ≤ 114$ square feet
The area $A$ of a rectangle is given as follows:
$A = L × W$
As per the question, the area, $A = L × W$ must not be exceeded by $114$ ft.
Then
$L × W ≤ 114$
Substitute $L = 25 – W$ from equation (i).
$(25-W) × W ≤ 114$
$25W - W^{2} ≤ 114$
$W^{2} – 25W + 114 ≤ 0$
$W^{2} – 19W - 6W + 114 ≤ 0$
$W (W – 19) -6(W – 19) ≤ 0$
$(W- 6) (W – 19) ≤ 0$
$W = 6$ ft or $19$ ft
Now, calculate $L$.
$L = 25 – W$
This implies
$L = 25 – 6 = 19$ ft
or
$L = 25 – 19 = 6$ ft
Hence, $L = 19$ ft or $6$ ft and $W = 6$ ft or $19$ ft.
Since $A ≤ 114$ sq ft, any length $L ≤ 19$ ft will be the possible length of a side.
However, the inequality involves (less than or equal to), we must also include the solution of $W^{2} – 25W + 114 ≤ 0$, namely $6$, $19$, in the solution set.
Therefore, we conclude that the possible sides (in feet) are in $(0, 6]$ or $[19, 25)$.
