2.7 Mathematical Models: Constructing Functions
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1 CHAPTER Functions and Their Graphs.7 Mathematical Models: Constructing Functions OBJECTIVE Construct and Analyze Functions Construct and Analyze Functions Real-world problems often result in mathematical models that involve functions.these functions need to be constructed or built based on the information given. In constructing functions, we must be able to translate the verbal description into the language of mathematics. We do this by assigning symbols to represent the independent and dependent variables and then finding the function or rule that relates these variables. EXAMPLE Area of a Rectangle with Fied Perimeter Figure 7 l w A w l The perimeter of a rectangle is 5 feet. Epress its area A as a function of the length l of a side. Consult Figure 7. If the length of the rectangle is l and if w is its width, then the sum of the lengths of the sides is the perimeter, 5. The area A is length times width, so The area A as a function of l is l + w + l + w = 5 l + w = 5 l + w = 5 w = 5 - l A = lw = l5 - l Al = l5 - l Note that we use the symbol A as the dependent variable and also as the name of the function that relates the length l to the area A. This double usage is common in applications and should cause no difficulties. EXAMPLE Economics: Demand Equations In economics, revenue R, in dollars, is defined as the amount of money received from the sale of a product and is equal to the unit selling price p, in dollars, of the product times the number of units actually sold. That is, R = p
2 SECTION.7 Mathematical Models: Constructing Functions In economics, the Law of Demand states that p and are related: As one increases, the other decreases. Suppose that p and are related by the following demand equation: p = - +, Epress the revenue R as a function of the number of units sold. Since R = p and p = - it follows that +, R = p = a - + b = - + NOW WORK PROBLEM. Figure 7 EXAMPLE y y (, ) d (, y) Figure 7 Finding the Distance from the Origin to a Point on a Graph Let P =, y be a point on the graph of y = -. (a) Epress the distance d from P to the origin O as a function of. (b) What is d if =? (c) What is d if =? (d) What is d if =? (e) Use a graphing utility to graph the function d = d, Ú. Rounded to two decimal places, find the value(s) of at which d has a local minimum. [This gives the point(s) on the graph of y = - closest to the origin.] (a) Figure 7 illustrates the graph. The distance d from P to O is Since P is a point on the graph of y = -, we substitute - for y.then We have epressed the distance d as a function of. (b) If =, the distance d is (c) If =, the distance d is (d) If =, d = - + y - = + y d = + - = - + the distance d is d = = d = - + = da b = B a b - a b + = A - + = (e) Figure 7 shows the graph of Y = - +. Using the MINIMUM feature on a graphing utility, we find that when L.7 the value of d is smallest.the local minimum is d L.87 rounded to two decimal places. [By symmetry, it follows that when L -.7 the value of d is also a local minimum.] NOW WORK PROBLEM 9.
3 CHAPTER Functions and Their Graphs EXAMPLE Filling a Swimming Pool A rectangular swimming pool meters long and meters wide is meters deep at one end and meter deep at the other. Figure 7 illustrates a cross-sectional view of the pool.water is being pumped into the pool to a height of meters at the deep end. Figure 7 m m L m m m (a) Find a function that epresses the volume V of water in the pool as a function of the height of the water at the deep end. (b) Find the volume when the height is meter. (c) Find the volume when the height is meters. (d) At what height is the volume cubic meters? cubic meters? (a) Let L denote the distance (in meters) measured at water level from the deep end to the short end. Notice that L and form the sides of a triangle that is similar to the triangle whose sides are meters by meters. This means L and are related by the equation L = or L =, The volume V of water in the pool at any time is Since V = cross-sectional triangular area width = a Lb cubic meters L =, we have V = a # # b = cubic meters (b) When the height of the water is meter, the volume V = V is V = # = (c) When the height of the water is meters, the volume V = V is V = # = = cubic meters cubic meters (d) By solving the equation we find that when L.77 meter, the volume is cubic meters. By solving the equation we find that when =, =, L.7 meters, the volume is cubic meters.
4 SECTION.7 Mathematical Models: Constructing Functions Figure 7 y (,) EXAMPLE 5 (, y) y 5 5 Area of a Rectangle A rectangle has one corner on the graph of y = 5 -, another at the origin, a third on the positive y-ais, and the fourth on the positive -ais. See Figure 7. (a) Epress the area A of the rectangle as a function of. (b) What is the domain of A? (c) Graph A = A. (d) For what value of is the area largest? (a) The area A of the rectangle is A = y, where y = 5 -. Substituting this epression for y, we obtain A = 5 - = 5 -. (b) Since represents a side of the rectangle, we have 7. In addition, the area must be positive, so y = 5-7, which implies that 6 5, so Combining these restrictions, we have the domain of A as 5 ƒ or, 5 using interval notation. (c) See Figure 75 for the graph of A = A. (d) Using MAXIMUM, we find that the area is a maimum of 8. at =.89, each rounded to two decimal places. See Figure 76. Figure 75 Figure NOW WORK PROBLEM 5. 5 EXAMPLE 6 Figure 77 Making a Playpen* A manufacturer of children s playpens makes a square model that can be opened at one corner and attached at right angles to a wall or, perhaps, the side of a house. If each side is feet in length, the open configuration doubles the available area in which the child can play from 9 square feet to 8 square feet. See Figure 77. Now suppose that we place hinges at the outer corners to allow for a configuration like the one shown in Figure 78. Figure 78 h 8 ft * Adapted from Proceedings, Summer Conference for College Teachers on Applied Mathematics (University of Missouri, Rolla), 97.
5 CHAPTER Functions and Their Graphs (a) Epress the area A of this configuration as a function of the distance between the two parallel sides. (b) Find the domain of A. (c) Find A if = 5. (d) Graph A = A. For what value of is the area largest? What is the maimum area? (a) Refer to Figure 78. The area A that we seek consists of the area of a rectangle (with width and length ) and the area of an isosceles triangle (with base and two equal sides of length ).The height h of the triangle may be found using the Pythagorean Theorem. h + a b = h = - a b = 9 - = 6 - h = 6 - The area A enclosed by the playpen is A = area of rectangle + area of triangle = + a 6 - b Figure 79 A = + Now the area A is epressed as a function of. (b) To find the domain of A, we note first that 7, since is a length. Also, the epression under the square root must be positive, so Combining these restrictions, we find that the domain of A is 6 6 6, or, 6 using interval notation. (c) If = 5, the area is A5 = L 9.5 square feet 6 If the width of the playpen is 5 feet, its area is 9.5 square feet. (d) See Figure 79. The maimum area is about 9.8 square feet, obtained when is about 5.58 feet.
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