Rectangles You have a flexible fence of length L = 13 meters. You want to use all of this fence to enclose a rectangular plot of land of at least 8 square meters in area. 1. Determine a function for the area of the plot of land in terms of either the width or the length.. Sketch a graph of the function and determine a reasonable domain and range. 3. Illustrate on the graph the values for which the area of the plot of land is at least 8 meters. 4. Solve problem 3 algebraically. 5. Describe the relationship between the graphical and algebraic solutions. 6. What dimensions will give a plot of land with the greatest area? Chapter 4: Quadratic Functions 133
Notes Materials: Graphing calculator Algebra II TEKS Focus: (A.6) Quadratic and square root functions. The student understands that quadratic functions can be represented in different ways and translates among their various representations. The student is expected to: (A) determine the reasonable domain and range values of quadratic functions, as well as interpret and determine the reasonableness of solutions to quadratic equations and inequalities. (B) relate representations of quadratic functions, such as algebraic, tabular, graphical, and verbal descriptions. Additional Algebra II TEKS: (A.8) Quadratic and square root functions. The student formulates equations and inequalities based on quadratic functions, uses a variety of methods to solve them, and analyzes the solutions in terms of the situation. The student is expected to: (A) analyze situations involving quadratic functions and formulate quadratic equations Scaffolding Questions: Draw a diagram of some possible rectangular plots with a perimeter of 13 meters. If the length of one side of the plot is 4 meters, and the perimeter is 13 meters, draw the plot. What is the length of the other side? What is the area of the plot? If the length of one side of the plot is x meters, and the perimeter is 13 meters, what is the length of the other side? What is the area? Graph the area of the plot in terms of x. Sample Solutions: 1. Since the perimeter is 13 meters, half the perimeter is 65 meters. If the length of one side of the plot is x meters, the length of the other side is 65 x meters, and the area is x (65 x ) square meters: A(x) = x (65 x). Here is a graph of A(x). It is an inverted parabola: The domain for this function is x 65, and the range is y 156.5. 134 Chapter 4: Quadratic Functions
3. The auxiliary lines drawn show that if the side length x is between about 17 meters and about 48 meters, the area will be 8 square meters or more. 4. To find these dimensions exactly, solve the following equation for x: A(x) = x (65 x) = 8 This is a quadratic equation in x, and we can write it as x 65x + 8 = Its solutions can be found by the quadratic formula: 65 1 x = ± 65 4( 8) 3. 5 ± 16 The solutions are approximately x = 16.5 meters and x = 48.5 meters. Since half the perimeter is 65 meters, the side length 16.5 meters of one side of the rectangle corresponds to a side length of 65 16.5 = 48.5 meters for the other side. Therefore both solutions correspond to a 16.5 by 48.5 rectangle. 5. To ask when the area is greater than or equal to 8 square meters, solve the equation x(65 x) = 8 and determine the values of x for which the area is equal to 8 square meters. On the graph these values are the intersection of the lines y = 8 and the parabola y = x(65 x). 6. Since the parabola that is the graph of A(x) is symmetric about its two zeros x = and x = 65, the largest value of the area will occur halfway in between these, at x = 3.5. The other side length is then 65 3.5 = 3.5 meters. In other words, the rectangle giving the largest area is a square. The area of this square is 3.5 = 1,56.5 square meters. or inequalities to solve problems. (B) analyze and interpret the solutions of quadratic equations using discriminants and solve quadratic equations using the quadratic formula. (C) compare and translate between algebraic and graphical solutions of quadratic equations. (D) solve quadratic equations and inequalities using graphs, tables, and algebraic methods. Connection to TAKS: Objective 1: The student will describe functional relationships in a variety of ways. Objective : The student will of the properties and attributes of functions. Objective 5: The student will of quadratic and other nonlinear functions. Objective 7: The student will of two- and three-dimensional representations of geometric relationships and shapes. Objective 1: The student will of the mathematical processes and tools used in problem solving. Chapter 4: Quadratic Functions 135
Here are diagrams showing the range of shapes of plots with perimeter 13 meters and area at least 8 square meters. 136 Chapter 4: Quadratic Functions
Extension Questions: Solve the problem generally. If you have a fence of fixed length P meters, and want the area of a rectangular plot enclosed by this fence to be at least A square meters, what are the possible dimensions of the plot? P Since the perimeter is P meters, half the perimeter is meters. If the length of one P side of the plot is x meters, the length of the other side is x meters, and the P area is x x square meters: P A( x) = x x The requirement that the area be at least A square meters leads to the inequality: P x x A This is a quadratic inequality in x, and we can write it as x P x + A Its solutions can be found by first solving the corresponding quadratic equation x P x + A = using the quadratic formula. Its solutions are x = P P + P P A 4 or 4A Therefore the smallest one side can be and still yield at least an area of A is P P - 4A P 1 P or - 4 4A What are the largest and smallest areas that can be enclosed in a rectangular plot with a fixed perimeter P? There is no smallest area, since by making the rectangle very long and thin, the area can be made as small as desired and still have perimeter P. Chapter 4: Quadratic Functions 137
The largest area will occur when the rectangle is a square with sides P 4. P 4. Its area is 138 Chapter 4: Quadratic Functions