Problem 14 - A Graphical Approach to College Algebra

Question

Solve each problem. Maximizing Area A homeowner has 80 feet of fence mo enclose a rectangular garden. What dimensions for the garden give the maximum area?

Step-by-Step Solution

Answer

The dimensions for maximum area are 20 feet by 20 feet.

Step 1: Define the Variables

Let's define the variables involved. Let \( x \) be the length of the garden and \( y \) be the width of the garden. We need to determine \( x \) and \( y \) such that the area is maximized.

Step 2: Set Up the Perimeter Equation

The perimeter of the rectangular garden is given by the formula \( 2x + 2y = 80 \) feet, since the entire fencing is used for the perimeter.

Step 3: Simplify the Perimeter Equation

Divide the entire perimeter equation \( 2x + 2y = 80 \) by 2, resulting in \( x + y = 40 \).

Step 4: Express One Variable in Terms of the Other

Solve for \( y \) in terms of \( x \): \( y = 40 - x \). This allows us to express the area \( A \) as a function of a single variable.

Step 5: Define the Area Function

The area \( A \) of the rectangle is given by \( A = x imes y \). Substitute \( y = 40 - x \) into this equation to get \( A = x(40 - x) = 40x - x^2 \).

Step 6: Find the Critical Points

To maximize the area, find the critical points by taking the derivative of the area function \( A(x) = 40x - x^2 \) with respect to \( x \). The derivative is \( A'(x) = 40 - 2x \). Set this equal to zero: \( 40 - 2x = 0 \).

Step 7: Solve for Optimal Length

Solve \( 40 - 2x = 0 \) to find the optimal value of \( x \). This gives \( x = 20 \).

Step 8: Determine the Corresponding Width

Using \( y = 40 - x \), substitute \( x = 20 \) to find \( y \): \( y = 40 - 20 = 20 \).

Step 9: Verify the Solution

Ensure that the calculated dimensions \( x = 20 \) and \( y = 20 \) satisfy the condition of the perimeter: \( 2(20) + 2(20) = 80 \) feet, which is correct.

Key Concepts

Rectangular Area Optimization

Rectangular area optimization is a fascinating problem commonly found in algebra, where the objective is to maximize or minimize the area of a rectangle under certain conditions. It's a practical application of mathematical concepts like equations and functions.br> For instance, if we want to maximize the area of a garden with a given amount of fencing, we need to find the dimensions that give us the largest possible area.br> Given the perimeter constraint, we first set up our equations based on the problem requirements. Here, if a fixed amount of fencing available is known, we derive a function for the area based on different possible lengths and widths.br> Optimizing the area involves expressing the area formula in terms of one variable, usually through substitution. This simplifies the process and leads us to use calculus for finding the optimal dimensions. Once these dimensions are identified, they give us the desired maximum area the rectangular space can take.

Perimeter Constraints

Perimeter constraints refer to the limits placed on the perimeter of a shape while solving optimization problems, such as those in rectangular area optimization. Here, the perimeter is the total length of all the sides.br>In the garden problem, the constraint is that the total fencing available is 80 feet.br> By setting up the perimeter equation as \(2x + 2y = 80\), where \(x\) is the length and \(y\) the width, we're linking these dimensions with a fixed constraint.br> Simplifying this equation helps in expressing one variable in terms of the other. That step is crucial because it enables us to introduce calculus concepts for optimization. For example, dividing by two, we get \(x + y = 40\), which helps us easily substitute later on to find the area as a function of \(x\).br> The constraints guide the solution path since any dimensions chosen must satisfy this fundamental equation.

Derivative Applications

The application of derivatives is vital in optimization problems, notably in finding maximum or minimum values of functions, like the area in algebra problems.br> Here, to maximize the area of the garden, we need to consider the area function, which after substitutions, becomes \(A(x) = 40x - x^2\). This is a quadratic function expressing the area concerning one dimension, \(x\).br> Calculus allows us to find the maximum point of this function by taking its derivative. The derivative \(A'(x) = 40 - 2x\) provides the rate of change of the area as \(x\) changes.br> Setting \(A'(x) = 0\) finds the critical points, which tells us where the function could reach a maximum or minimum. Solving \(40 - 2x = 0\) gives us \(x = 20\), pinpointing where the area is optimized. Substituting back gives us the corresponding dimension \(y = 20\), confirming that both dimensions ensure the maximum area under set constraints.br> Thus, derivatives are powerful tools in transforming word problems into numerical solutions, especially in optimizing dimensions or other variables under given rules.