Maximizing an Area

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?

Figure 53 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 \([0,2000]\).

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.