Problem 16 - Beginning and Intermediate Algebra

Question

Solve. A farmer wants to enclose a rectangular area with \(120 \mathrm{ft}\) of fencing. One side is a river and will not require a fence. What is the maximum area that can be enclosed?

Step-by-Step Solution

Answer

The maximum area that can be enclosed by 120ft of fencing is 1800 square feet, with the dimensions 60ft by 30ft.

Step 1: Define the variables and Area function

Let's define \(x\) as the length parallel to the river, and \(y\) as the length perpendicular to the river. The fencing will be distributed as follows: \(x\) feet will cover the length parallel to the river, and \(2y\) feet will cover the two perpendicular lengths. Therefore, we can write the constraints of the problem as: \(x + 2y = 120\) Next, let's define the area function. Since the enclosed area is a rectangle, its area \(A\) can be found by multiplying the length and the width: \(A(x, y) = xy\)

Step 2: Solve the constraint equation for y

To express the Area function \(A(x, y)\) in terms of just one variable, we will solve the constraint equation from Step 1 for \(y\): \(y = \frac{120 - x}{2}\)

Step 3: Substitute y in the Area function

Replace the \(y\) with the expression we found in Step 2: \(A(x) = x(\frac{120 - x}{2})\) Now we have the area function in terms of one variable: \(A(x) = \frac{1}{2}x(120 - x)\)

Step 4: Differentiate the Area function

To find the maximum area, we will differentiate the area function with respect to \(x\): \(A'(x) = \frac{d}{dx}(\frac{1}{2}x(120 - x))\) To differentiate the function, we will use the product rule: \(A'(x) = \frac{1}{2}(120x - x^2)'\) \(A'(x) = 60 - x\)

Step 5: Find the critical points

Equate the derivative to zero to find the critical points: \(60 - x = 0\) Solve for \(x\): \(x = 60\)

Step 6: Find the maximum enclosed area

Now that we have found the critical point \(x = 60\), we can substitute this value back into the constraint equation to find the corresponding value of \(y\): \(y = \frac{120 - 60}{2}\) \(y = 30\) Finally, we can find the maximum enclosed area by substituting these values into the area function: \(A(60) = 60 \times 30\) \(A(60) = 1800\) The maximum area that can be enclosed by 120ft of fencing is 1800 square feet, with the dimensions 60ft by 30ft.

Key Concepts

Rectangular area problem

In the rectangular area problem, the goal is to find the maximum or minimum area of a rectangle under certain constraints. In our example, a farmer wants to enclose a rectangular area along a river using fencing. Only three sides require fencing since the fourth side is the river itself. This is a common type of real-world problem that can be solved using optimization methods in calculus.

To approach problems like this, it's important to define what you know and what you want to find out. For instance, here:

• Known: 120 feet of fencing available
• Unknown: Dimensions of the rectangle that would result in maximum area
• Constraint: Fencing covers only three sides, no fence is needed along the river

Start by setting up an equation using the constraint (in this case, total length of fencing). Then, express the area as a function of one variable, make sure to clearly express each step, as these problems often involve restating the problem through algebraic manipulation.

The process of solving a rectangular area problem not only helps in finding the solution but improves skills in transforming practical questions into mathematical solutions.

Critical points in calculus

Critical points are values in a function where its derivative hits zero or does not exist. These points are crucial when finding maximum or minimum values of a function, especially in optimization problems like the rectangular area problem.

By setting the derivative of an area function to zero, you can determine where the area might reach its maximum. This is done because, at critical points, the slope of the tangent to the curve is zero, indicating a potential peak (maximum) or bottom (minimum).

In our exercise solution, the area function was differentiated, resulting in an equation which was then set to zero to find the critical point:

• Differentiate area function with respect to the length ( ).
• Find where this derivative equals zero, that is, " = 60" in the solution.

This critical point tells us that a rectangle with a side length of 60 having a maximum area for the constraints given.

Understanding critical points helps in solving not just calculus problems, but any situation requiring determining optimal conditions.

Product rule differentiation

Product rule differentiation is a technique used when differentiating functions that are the product of two or more functions. In calculus, this rule is critical when solving problems like optimizing rectangular areas.

The product rule states: if you have a function that is the product of two functions, say \( u(x) \) and \( v(x) \), the derivative is given by:

• \( (uv)' = u'v + uv' \)

In the context of our area function \( A(x) = \frac{1}{2}x(120 - x) \), the product rule was used to find its derivative:

• Identify parts: \( u(x) = x \) and \( v(x) = \frac{120 - x}{2} \)
• Differentiate each: \( u'(x) = 1 \) and \( v'(x) = -\frac{1}{2} \)
• Apply the rule: calculate \( A'(x) \) using \( (uv)' \)

By using the product rule, we achieve the derivative necessary to find critical points. Employing such rules simplifies the process of handling complex functions and is vital in both theoretical and applied calculus problems.