Respuesta :
Let the length of the rectangle be "x" and the width be "y".
We know the formula for the area of a rectangle:
[tex]\begin{gathered} A=\text{length}\times\text{width} \\ A=xy \end{gathered}[/tex]We know the formula for the perimeter of a rectangle:
[tex]\begin{gathered} \text{Perimeter}=2(length+width) \\ 56=2(x+y) \\ \frac{56}{2}=x+y \\ x+y=28 \end{gathered}[/tex]Let's solve for "x" in the perimeter equation to get:
[tex]x=28-y[/tex]Now, we will substiute this into the Area equation. Shown below:
[tex]\begin{gathered} A=xy \\ A=(28-y)y \\ A=28y-y^2 \end{gathered}[/tex]To find the maximum area, we need to differentiate "A" equation and set it equal to 0 to find the value of "y" for which "A" is maximum. The steps are shown below:
[tex]\begin{gathered} A=28y-y^2 \\ A^{\prime}=28-2y \\ 28-2y=0 \\ 2y=28 \\ y=\frac{28}{2} \\ y=14 \end{gathered}[/tex]The corresponding "x" value is:
[tex]\begin{gathered} x=28-y \\ x=28-14 \\ x=14 \end{gathered}[/tex]Thus, the rectangle with the maximum area is the one with dimensions:
[tex]\begin{gathered} \text{Length}=14yd \\ \text{Width}=14yd \end{gathered}[/tex]