Question

Mr. Creenthumb wishes to mark out a rectangular flower bed, using a wall of his house as one side of the rectangle. The other three sides are to be marked by wire netting, of which he has only 64 ft available. What are the length \(L\) and width \(W\) of the rectangle that would give him the largest possible planting area? How do you make sure that your answer gives the largest, not the smallest area?

Short answer

The optimal dimensions are \( L = 16 \) and \( W = 32 \), giving the largest area.

Step-by-step solution

Identifying the Variables

In this problem, we need to maximize the area of a rectangular flower bed. Let's denote the width of the rectangle parallel to the house wall as \( W \) (the part not using netting) and the length perpendicular to the wall as \( L \). The objective is to find \( L \) and \( W \) such that the area is maximized.

Setting up the Perimeter Equation

Since the wall of the house forms one side of the rectangle, the wire netting will cover the two lengths \( L \) and one width \( W \). Thus, the equation for the wire netting is: \[ 2L + W = 64 \] which accounts for two lengths (both perpendicular to the house) and one width parallel to the house.

Expressing Area Function

The area \( A \) of the rectangle is given by the formula: \[ A = L \times W \] Since we want to maximize the area, we need to express it in terms of one variable. Substitute \( W \) from the perimeter equation: \[ W = 64 - 2L \] Now, the area in terms of \( L \) is: \[ A(L) = L \times (64 - 2L) = 64L - 2L^2 \]

Finding the Maximum Area

To find the maximum area, we need to differentiate the area function \( A(L) \) and find its critical points:\[ \frac{dA}{dL} = 64 - 4L \]Set the derivative equal to zero to find the critical points:\[ 64 - 4L = 0 \]\[ L = 16 \]

Determining the Optimal Width

With \( L = 16 \), use the previously derived equation to determine \( W \): \[ W = 64 - 2 \times 16 = 32 \]

Verifying Maximum Area

Since the area function \( A(L) = 64L - 2L^2 \) is a quadratic function opening downwards (as indicated by the negative coefficient of \( L^2 \)), its vertex represents the maximum point. Therefore, \( L = 16 \) and \( W = 32 \) provide the rectangle with the largest area.

Key concepts

Quadratic Functions

Quadratic functions are expressions of the form \( ax^2 + bx + c \), where \(a\), \(b\), and \(c\) are constants. These functions graph as parabolas, which can open upwards or downwards. The key feature of a quadratic function is its vertex. For a parabola that opens downwards, the vertex is at the top and represents the maximum point of the function. In the given problem, the area equation \( A(L) = 64L - 2L^2 \) is a quadratic function. The coefficient of \( L^2 \) is negative, meaning the parabola opens downwards, and the maximum area is found at the vertex.

Critical Points

Critical points occur where the derivative of a function equals zero or is undefined. These points are vital in determining maximum and minimum values of functions. In the exercise, finding critical points helps us to identify the maximum area of the rectangular garden using differentiation. We set the derivative of the area function \( \frac{dA}{dL} = 64 - 4L \) equal to zero to find \( L = 16 \). These critical points show where the slope of the function is flat, indicating potential maxima or minima. In this scenario, the critical point at \( L = 16 \) is a maximum because of the downward-opening parabola.

Differentiation

Differentiation is a mathematical process used to find the rate at which a function is changing at any given point. It's essential for solving optimization problems, like the maximum area problem here. By differentiating the area function \( A(L) = 64L - 2L^2 \), we obtain \( \frac{dA}{dL} = 64 - 4L \). This derivative tells us how the area changes concerning changes in \( L \). We then set this derivative equal to zero to find where the maximum or minimum points are, helping us pinpoint the optimal dimensions for the largest planting area.

Maximum Area Problems

Maximum area problems involve optimizing the dimensions of a shape to get the largest possible area within given constraints. For Mr. Creenthumb's flower bed, we had to maximize the area while only having a fixed amount of wire netting for the perimeter. The strategy involves writing equations for the area and the perimeter and expressing the area in terms of a single variable. By transforming the perimeter constraint \( 2L + W = 64 \) to express \( W \) in terms of \( L \), and then substituting it into the area formula, we get a quadratic function \( A(L) = 64L - 2L^2 \). Solving this using differentiation allows us to find the length and width that deliver the largest area under the given perimeter constraint.