The farmer would like to eclose a rectangle
First we know that the perimeter of the rectangle is
[tex]P=2x+2y[/tex]We also know that we have 36 ft of fence, that is we only can enclose a rectangle of 36 ft of perimeter. Then
[tex]36=2x+2y[/tex]From this we can find y
[tex]\begin{gathered} 36=2x+2y \\ 36-2x=2y \\ y=\frac{36-2x}{2} \\ y=18-x \end{gathered}[/tex]The area of the rectangle is
[tex]A=xy[/tex]But we know the value of y, plugging this value into the last equation we have that
[tex]A=x(18-x)[/tex]To find three possible values for the area we only have to give values to x. This values have to be positive (since we can't have a negative lenght). We also notice that the value can't exceed 18 since that would mean a zero area. With those point in consideration we choose three values between zero and 18.
If x=3, then the area is
[tex]\begin{gathered} A=3(18-3) \\ =3(15) \\ =45 \end{gathered}[/tex]If x=9, then the area is
[tex]\begin{gathered} A=9(18-9) \\ =9(9) \\ =81 \end{gathered}[/tex]if x=15, then the area is
[tex]\begin{gathered} A=15(18-15) \\ =15(3) \\ =45 \end{gathered}[/tex]Then we have three possible areas for the rectangle.
The maximum value for the area is 81, and we see that because the equation for the area is a parabola that opens down with vertex in the point (9,81)
