The perimeter of the rectangular plot of land is given by the expression below
![P=2x+2y](https://tex.z-dn.net/?f=P%3D2x%2B2y)
On the other hand, since the available money to buy fence is D dollars,
![\begin{gathered} D=4(2x)+3(2y) \\ \Rightarrow D=8x+6y \\ D\rightarrow\text{ constant} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20D%3D4%282x%29%2B3%282y%29%20%5C%5C%20%5CRightarrow%20D%3D8x%2B6y%20%5C%5C%20D%5Crightarrow%5Ctext%7B%20constant%7D%20%5Cend%7Bgathered%7D)
Furthermore, the area of the enclosed land is given by
![A=xy](https://tex.z-dn.net/?f=A%3Dxy)
Solving the second equation for x,
![\begin{gathered} D=8x+6y \\ \Rightarrow x=\frac{D-6y}{8} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20D%3D8x%2B6y%20%5C%5C%20%5CRightarrow%20x%3D%5Cfrac%7BD-6y%7D%7B8%7D%20%5Cend%7Bgathered%7D)
Substituting into the equation for the area,
![\begin{gathered} A=(\frac{D-6y}{8})y \\ \Rightarrow A=\frac{D}{8}y-\frac{3}{4}y^2 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20A%3D%28%5Cfrac%7BD-6y%7D%7B8%7D%29y%20%5C%5C%20%5CRightarrow%20A%3D%5Cfrac%7BD%7D%7B8%7Dy-%5Cfrac%7B3%7D%7B4%7Dy%5E2%20%5Cend%7Bgathered%7D)
To find the maximum possible area, solve A'(y)=0, as shown below
![\begin{gathered} A^{\prime}(y)=0 \\ \Rightarrow\frac{D}{8}-\frac{3}{2}y=0 \\ \Rightarrow\frac{3}{2}y=\frac{D}{8} \\ \Rightarrow y=\frac{D}{12} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20A%5E%7B%5Cprime%7D%28y%29%3D0%20%5C%5C%20%5CRightarrow%5Cfrac%7BD%7D%7B8%7D-%5Cfrac%7B3%7D%7B2%7Dy%3D0%20%5C%5C%20%5CRightarrow%5Cfrac%7B3%7D%7B2%7Dy%3D%5Cfrac%7BD%7D%7B8%7D%20%5C%5C%20%5CRightarrow%20y%3D%5Cfrac%7BD%7D%7B12%7D%20%5Cend%7Bgathered%7D)
Therefore, the corresponding value of x is
![\begin{gathered} y=\frac{D}{12} \\ \Rightarrow x=\frac{D-6(\frac{D}{12})}{8}=\frac{D-\frac{D}{2}}{8}=\frac{D}{16} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20y%3D%5Cfrac%7BD%7D%7B12%7D%20%5C%5C%20%5CRightarrow%20x%3D%5Cfrac%7BD-6%28%5Cfrac%7BD%7D%7B12%7D%29%7D%7B8%7D%3D%5Cfrac%7BD-%5Cfrac%7BD%7D%7B2%7D%7D%7B8%7D%3D%5Cfrac%7BD%7D%7B16%7D%20%5Cend%7Bgathered%7D)
<h2>Thus, the dimensions of the fence that maximize the area are x=D/16 and y=D/12.</h2><h2>As for the used money,</h2>
![\begin{gathered} top,bottom:\frac{8D}{16}=\frac{D}{2} \\ Sides:\frac{6D}{12}=\frac{D}{2} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20top%2Cbottom%3A%5Cfrac%7B8D%7D%7B16%7D%3D%5Cfrac%7BD%7D%7B2%7D%20%5C%5C%20Sides%3A%5Cfrac%7B6D%7D%7B12%7D%3D%5Cfrac%7BD%7D%7B2%7D%20%5Cend%7Bgathered%7D)
<h2>Half the money was used for the top and the bottom, while the other half was used for the sides.</h2>