A farmer with $4000 \,{\rm m}$ of fencing wants to enclose a rectangular plot that borders a straight river, as shown in Figure 53. If the farmer does not fence the side along the river, what is the largest rectangular area that can be enclosed?

Area $A = xy$

Solution

Step 1 The quantity to be maximized is the area; we denote it by $A$.

Step 2 We denote the dimensions of the rectangle by $x$ and $y$ (both in meters), with the length of the side $y$ parallel to the river. The area is $A=xy$. Because there are $4000 \,{\rm m}$ of fence available, the variables $x$ and $y$ are related by the equation $$\begin{eqnarray*} x+y+x &=&4000 \\[3pt] y &=&4000-2x \end{eqnarray*}$$

Step 3 Now we express the area $A$ as a function of $x$. $$A=A(x) =x(4000-2x)=4000x-2x^{2}\qquad \color{#0066A7}{{\hbox{\(A=xy,\quad y=4000-2x\)}}}$$

The domain of $A$ is the closed interval $[0,2000]$.

Step 4 To find the number $x$ that maximizes $A(x) ,$ we differentiate $A$ with respect to $x$ and find the critical numbers: $${A}^{\prime }(x)=4000-4x=0$$

The critical number is $x=1000$. The maximum value of $A$ occurs either at the critical number or at an endpoint of the interval $[0,2000]$. $$A({1000})=2{,}000{,}000\qquad {A(0)=0} \qquad {A({2000})}=0$$

The maximum value is $A(1000)=2{,}000{,}000\ {\rm{m}}^{2},$ which results from using a rectangular plot that measures $2000\ {\rm{m}}$ along the side parallel to the river and $1000\ {\rm{m}}$ along each of the other two sides.