Problem 47

Question

You have 120 feet of fencing to enclose a rectangular plot that borders on a river. If you do not fence the side along the river, find the length and width of the plot that will maximize the area. What is the largest area that can be enclosed?

Step-by-Step Solution

Verified
Answer
The dimensions that maximize the area are 30 feet by 60 feet, and the maximum area is 1800 square feet.
1Step 1: Setting up the Problem
Let \(x\) denote the length of the fence perpendicular to the river, and \(y\) denote the length of the fence along the river. We know that \(2x + y = 120\) since you have 120 feet of fencing. This is our constraint equation.
2Step 2: Express the Area as a Function of a Single Variable
The area of the rectangular plot can be expressed as \(A = xy\). Considering our constraint equation, we can solve for \(y\), such that \(y = 120 - 2x\). Substituting this equation into \(A = xy\), we get \(A = x(120 - 2x)\), or simplifying, \(A = 120x - 2x^2\).
3Step 3: Find the Maximum Area Using Differentiation
To maximize the area, we take the derivative of the area function \(A(x) = 120x - 2x^2\) with respect to \(x\), set the resulting equation equal to zero, and solve for \(x\). The derivative \(A'(x) = 120 - 4x\). Setting this to zero, yields the equation \(120 - 4x = 0\), thus \(x = 30\). To ensure this value provides a maximum area, we can examine the second derivative of \(A(x)\), it is \(A''(x) = -4\), which is always negative. This suggests that the area \(A\) is at a maximum when \(x = 30\).
4Step 4: Find the Dimensions that Maximize the Area
Substituting \(x = 30\) into the constraint equation \(2x + y = 120\) yields \(y = 120 - 2*30 = 60\). Thus, the length and width of the plot that will maximize the area are 30 feet (perpendicular to the river) and 60 feet (along the river) respectively.
5Step 5: Calculate the Maximum Area
The maximum area can be calculated by substituting the optimal dimensions into the area function \(A = xy\). This gives \(A = 30*60 = 1800\) square feet.