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?
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 \([0,2000]\).
Since \(A=A(x)\) is continuous on the closed interval [0, 2000], it will have an absolute maximum.
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.