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Use a table to find the maximum area for a rectangle with a perimeter of 88 feet.

Complete the table:
\begin{tabular}{|c|c|}
\hline
Dimensions & Area \\
\hline
[tex]$28 ft$[/tex] by [tex]$16 ft$[/tex] & [tex]$448 ft^2$[/tex] \\
\hline
[tex]$26 ft$[/tex] by [tex]$18 ft$[/tex] & [tex]$468 ft^2$[/tex] \\
\hline
[tex]$24 ft$[/tex] by [tex]$20 ft$[/tex] & [tex]$480 ft^2$[/tex] \\
\hline
[tex]$22 ft$[/tex] by [tex]$22 ft$[/tex] & [tex]$484 ft^2$[/tex] \\
\hline
\end{tabular}

The maximum area is [tex]$\square$[/tex] [tex]$ft^2$[/tex] for a rectangle with length [tex]$\square$[/tex] [tex]$ft$[/tex] and width [tex]$\square$[/tex] [tex]$ft$[/tex].

Sagot :

To find the maximum area for a rectangle with a perimeter of 88 feet, we can analyze the given table which has calculated areas based on different dimensions that satisfy the perimeter condition. The perimeter of a rectangle is given by [tex]\( P = 2(L + W) \)[/tex], where [tex]\( L \)[/tex] is the length and [tex]\( W \)[/tex] is the width. The perimeter is 88 feet, so:

[tex]\[ 2(L + W) = 88 \][/tex]
[tex]\[ L + W = 44 \][/tex]

The table below shows the dimensions and their corresponding areas for a few variations:

[tex]\[ \begin{tabular}{|c|c|} \hline \textbf{Dimensions} & \textbf{Area} \\ \hline 28 ft by 16 ft & 448 ft² \\ \hline 26 ft by 18 ft & 468 ft² \\ \hline 24 ft by 20 ft & 480 ft² \\ \hline 22 ft by 22 ft & 484 ft² \\ \hline \end{tabular} \][/tex]

From the table:

- The area for the dimensions 28 ft by 16 ft is 448 ft².
- The area for the dimensions 26 ft by 18 ft is 468 ft².
- The area for the dimensions 24 ft by 20 ft is 480 ft².
- The area for the dimensions 22 ft by 22 ft is 484 ft².

The maximum area is [tex]\( \boxed{484} \)[/tex] ft² for a rectangle with a length of [tex]\( \boxed{22} \)[/tex] ft and a width of [tex]\( \boxed{22} \)[/tex] ft.