The perimeter of the rectangular plot of land is given by the expression below
[tex]P=2x+2y[/tex]
On the other hand, since the available money to buy fence is D dollars,
[tex]\begin{gathered} D=4(2x)+3(2y) \\ \Rightarrow D=8x+6y \\ D\rightarrow\text{ constant} \end{gathered}[/tex]
Furthermore, the area of the enclosed land is given by
[tex]A=xy[/tex]
Solving the second equation for x,
[tex]\begin{gathered} D=8x+6y \\ \Rightarrow x=\frac{D-6y}{8} \end{gathered}[/tex]
Substituting into the equation for the area,
[tex]\begin{gathered} A=(\frac{D-6y}{8})y \\ \Rightarrow A=\frac{D}{8}y-\frac{3}{4}y^2 \end{gathered}[/tex]
To find the maximum possible area, solve A'(y)=0, as shown below
[tex]\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}[/tex]
Therefore, the corresponding value of x is
[tex]\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}[/tex]
Thus, the dimensions of the fence that maximize the area are x=D/16 and y=D/12.
As for the used money,
[tex]\begin{gathered} top,bottom:\frac{8D}{16}=\frac{D}{2} \\ Sides:\frac{6D}{12}=\frac{D}{2} \end{gathered}[/tex]
Half the money was used for the top and the bottom, while the other half was used for the sides.