You have D dollars to buy fence to enclose a rectangular plot of land (see figure at right). The fence for the top and bottom costs $4 per foot and for the sides it costs $3 per foot . Find the dimensions of the plot with the largest area. For this largest plot , how much money was used for the top and bottom, x, and for the sides, ?

You have D dollars to buy fence to enclose a rectangular plot of land see figure at right The fence for the top and bottom costs 4 per foot and for the sides it class=

Respuesta :

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.