Let the length of the rectangle be l and width of rectangle be w, then
[tex]\begin{gathered} \text{length}=l \\ \text{width}=w \\ \text{Perimeter}=2(l+w) \end{gathered}[/tex]Given that the perimeter of the rectangle is 56, then
[tex]\begin{gathered} 2(l+w)=52 \\ l+w=\frac{52}{2} \\ l+w=26 \end{gathered}[/tex]The area of the rectangle would be
[tex]A_{\text{rectangle}}=l\times w[/tex]From the derived perimeter of the rectangle, we can make l, the subject as shown below
[tex]\begin{gathered} l+w=26 \\ l=26-w \end{gathered}[/tex]Substituting for l in the area would give us
[tex]\begin{gathered} A_{\text{rectangle}}=l\times w;l=26-w \\ A_{\text{rectangle}}=(26-w)\times w \\ A_{\text{rectangle}}=26w-w^2 \end{gathered}[/tex]in other to get the maximum area, let complete the squares of the area as shown below
[tex]\begin{gathered} 26w-w^2=-w^2+26w \\ -w^2+26w=-1(w^2-26w) \\ \text{add square of half coefficient of w,} \\ \text{coefficient of w=-26} \\ \text{half coefficient of w=}\frac{1}{2}\times-26=-13 \\ squareofhalfcoefficientofw=(-13)^2=169 \end{gathered}[/tex]The area would become:
[tex]A_{\text{rectangle}}=-1(w-13)^2+169[/tex]The graph of the area is as shown below
From the above, the maximum area is 169 square unit.
The maximum area occur when the width, w = 13
The length would give us
[tex]\begin{gathered} l=26-w \\ l=26-13 \\ l=13 \end{gathered}[/tex]Since the length and width is the same, then the rectangle is a square.
Hence, the dimensions of the rectangle (i.e. square) with maximum area is (13 by 13)
