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1.8: Functions and Mathematical Models

Difficulty Level: At Grade Created by: CK-12

Learning objectives

• Categorize some realistic situations in terms of function families.
• Write a function to represent a situation.
• Use a function to determine key aspects of a situation.

Introduction

Throughout this chapter we have examined different kinds of functions and their behavior, and we have used functions to represent realistic situations. When we use a function to help us understand phenomena such as how to maximize the volume of a container or to minimize its surface area, we are engaging in mathematical modeling. In reality, scientists and social scientists use mathematical models to understand a wide variety of quantifiable phenomena, from the workings of subatomic particles, to how people will function in the economy.

In this lesson we will revisit some of the examples we have seen in previous lessons, in an effort to categorize models according to function families. We will look at several examples of models in depth, specifically in terms of how we can use a graphing calculator to help us analyze models.

Linear models

The very first example of a function in this chapter was a linear model. The equation y=3x was used to represent how much money you would bring in if you sold x boxes of cookies for $3 per box. Many situations can be modeled with linear functions. The key idea is that some quantity in the situation has a constant rate of change. In the cookie-selling example, every box costs$3.00. Therefore the profits increase at a constant rate.

The cookie-selling model is an equation of the form y=mx . The function necessarily contains the point (0,0) : if we don’t sell any cookies, we don’t bring in any money!

Other models will be of the form y=mx+b . The constant b is the y-intercept of the function, and represents the value of the function when x is zero. For example, consider a situation in which you plan to save money at a constant rate of $20 per week. If you begin to save money after receiving a gift of$100, you can express the amount you have saved as a function of time: S(t)=20t+100 , where t represents the number of weeks you have been saving. The function is linear because of the constant rate of change, that is, the constant savings of $20 per week. Notice that in both of these examples we will only consider these functions for x values ≥ 0. In the first example, x represents the numbers of boxes of cookies, which cannot be negative. In the second example, x represents the number of weeks you have been saving money. In theory we could extend this situation back in time, but the given information does not indicate that the model would make sense. This is the case because you received$100 as a gift at a particular point in time. You didn’t save that $100 at$20 per week.

Both of these examples also are linear functions with positive slope. In both situations, the function increases at a constant, or steady rate. We could also use a linear function to model a situation of constant decrease.

In sum, linear functions are used to model a situation of constant change, either increase or decrease. Next we will consider functions that can be used to model other kinds of situations.

In lesson 2 we saw several quadratic functions that were used as models. For example, the function $A(x)=50x-\frac{x^2}{2}$ was used to represent the area of a rectangular plot of land enclosed on three sides by 100 feet of fence. We used this model to identify the maximum possible area for the plot of land.

Because the graph is a parabola and the coefficient of x2 is negative, it represents an x-axis reflection (i.e., it is “upside-down”). Thus we know that the vertex will be a global maximum. This maximum point represents the maximum possible area for the enclosure. In the homework problems for lesson 2, you were asked to represent other situations using quadratic models. These situations were all related to the area and perimeter of rectangles. However, other kinds of phenomena can also be modeled with quadratic functions.

Consider for example a situation in which a ball is tossed into the air. The ball will travel up, and then it will travel down until it hits the ground. How high will the ball go? When will it reach the ground? This kind of situation can be modeled by a function of the form h(t)=-16t2+v0t+h0 . The variable t represents the time since the ball was thrown. The coefficient v0 represents the initial velocity of the ball, and the constant h0 represents the initial height of the ball. The constant -16 come from the force of gravity which pulls the ball down, which is why it is negative. The following example shows a specific situation of this form.

Example 1: You are standing on the roof of a building that is 20 feet above the ground. You toss a ball into the air with an initial velocity of 40 ft/sec, so that it will land on the ground, not on the roof. How high will the ball go, and when will it reach its maximum height? When will the ball hit the ground?

Solution: First we need to write a function to model the situation. Using the general form of the equation given above, we can write the function h(t)=-16t2+40t+20 , where h(t) represents the height above the ground..

To answer the first question, we need to examine the graph of the function. If you graph this equation on your calculator, you will need to determine a good viewing window. One way to start to determine a good window is to take into account the y-intercept of the function. In this case, the y-intercept is (0, 20). Also think about what kind of function this is: a parabola, facing downwards. This fact should lead you to think that we need to look at y-values well above 20. As noted earlier in the chapter, it is often useful to look at a table of values. Using the “ask” capability of the table, if you input x values of 1, 2, and 3, you will see that the function goes up to 44 at x = 1. The maximum value is most likely somewhere near x = 1. Press WINDOW, and set xmin = 0, xmax = 3. Then set Ymin =0 (or a little less, if you want to see the y-axis). Ymax should be no less than 44, though you may want to make it larger, such as 50 or more, just to be sure that you can see the vertex. Once you have set the window, press GRAPH.

Now you should see the parabola. To identify the coordinates of the vertex, you can use the MAX function in the CALC menu, introduced earlier in the chapter. Remember that the calculator will ask you to input a left bound, a right bound, and a guess for the maximum. If you use the MAX function, you should find that the coordinates of the vertex are (1.25, 45). This means that 1.25 seconds after the ball is thrown into the air, it reaches a maximum height of 45 feet.

Now, to answer the second part of the question, we need to determine when the height of the ball is 0. Graphically, we are looking for the x-intercept of the parabola. If you return to the GRAPH screen, you should see that the x-intercept is around 3. If we want to determine the exact value, or at least a good approximation of the x-intercept, we can use the ZERO function. Press 2nd TRACE to get the CALC menu, and choose option 2, ZERO. Like the MAX function, you need to input a left bound, a right bound, and a guess, although the guess is optional – just press ENTER. (Note that the calculator works this way because it is asking you to identify which x-intercept to calculate. The parabola has two x-intercepts, and other functions may have more.) Using the ZERO function, you should find that the x-intercept is approximately 2.93. This means that the ball reaches the ground in just under 3 seconds.

Like the area situations, the falling object is represented by a quadratic function that has a maximum. A quadratic function whose vertex is a global minimum could be used to model a situation in which some quantity decreases, and then increases. In general, polynomial functions can be used to model situations in which a quantity increases and decreases, or the rate of change changes, depending on the value of x. Next we will examine cubic functions that can be used to represent such situations.

Cubic models

In lesson 3 we analyzed a situation in which a rectangular piece of cardboard was folded into a box. The folding was possible because we cut squares out of the four corners of the cardboard. We used the function v(x)=(12-2x)(8-2x)x to represent the volume of the box as a function of x, the side-length of the squares cut out of the corners. If we multiply out the factors of this function we can verify that this is a cubic function:

v(x)=(12-2x)(8-2x)x=4x3-40x3+96x

We used this function to find the maximum possible volume of the box. We can also analyze the graph to understand how the volume changes as a function of x.

When we analyzed this function in lesson 3, we only looked at the portion of the graph that looks “parabolic”. We did this because the function ceases to model the situation if x is more than 4. If we cut out {4x4} squares, we would cut out one entire side of the cardboard rectangle, and we would not be able to make a box. Focusing then on the interval (0, 4), we can see that the volume of the box increases, and then decreases. If we wanted to know the volume of a box with particular dimensions, we can trace on the graph, input values into the table, or take advantage of the graph being in trace mode. That is, if you press GRAPH to view the graph, and then press TRACE, you can input x values. For example, say that you wanted to cut out squares of side-length 2.5. Press TRACE, then press 2.5, then press ENTER. At the bottom of the screen you will see x = 2.5 and y = 52.5. This tells you that the volume of the box will be 52.5in3.

You can also determine the value of x, given a specific volume. Say that you want the box to have a volume of 50 in3. One way to determine the value of x is to graph the constant function y = 50, and find the point where the volume function intersects y = 50. Press Y= and enter 50 into Y2. Now press GRAPH. You should see the horizontal line y = 50 intersecting the volume function in several places. We are interested in the two intersection points in the interval (0, 4).

To find a good approximation of an intersection point, trace close to one of the two points. If you trace close to the first point, you will see that it is around x = 0.8. Now press 2nd, CALC, and choose option 5, INTERSECT. The calculator will send you back to the graph screen, and ask you to choose the first curve. (The calculator does this in case you have more than two functions graphed at the same time.) You should see the cubic equation at the top of the screen. Press ENTER, and the calculator will ask you for the second curve. You should see y = 50 at the top of the screen. Press ENTER, and then enter a guess. (If you have already traced close to an intersection point, and you only have two functions graphed, you can simply press ENTER three times.) You should now see the coordinates of the intersection point at the bottom of the screen: x is approximately 0.723. (You can use the same process to estimate the other intersection point.)

A cubic function can be used to model situations that involve volume, but they can also be used to model situations that follow particular growth patterns. For example, a cubic function can be used the model cost of producing commercial goods. In the graph below, the x-axis represents hundreds of goods produced and the y-axis represents cost of production.

If the cost to manufacture each item was fixed, then the model would be linear. However, if we produce a larger number of items, we may be able to reduce the cost per item. For example, in the graph above, notice that the slope of the graph flattens out from about 1.5 to 4. If this graph is modeling production of x goods (in hundreds) then we can say that the manufacturing costs are increasing less quickly in this interval because we are likely to be spending less per term, when we produce more than 150 items at a time. The costs increase faster again when we are producing more than 400 items, perhaps because we might not have the resources to produce that many items, or more people or overtime would be required. In the review questions, you will examine a specific example of a cost function.

Now we turn to other kinds of models.

Other models

Besides linear, quadratic, and cubic functions, in this chapter we have seen other functions that can also be used to model situations. For example, the equation S(x)= (48/x)+2x2 was used in lesson 2 to model the surface area of a box. In a review question in lesson 3, you wrote an equation to model the area of a rectangle inscribed under a semi-circle. This equation belonged to the square root family. In lesson 5 we examined piece-wise functions, which can be used to describe situations in which quantitative relationships are different in different intervals within the domain of the function. For example, consider a situation in which a wireless provider offers customers a monthly plan that costs $50, but then charges$0.40 cents per minute for every minute over 1000 included day time minutes. We can model the monthly cost, C, of the plan as a function of m, the number of day time minutes you use:

$C(m) = \begin{cases} 50, & {m \le 1000} \\ 50+0.40(m-1000), & {m > 1000} \end{cases}$

This function is comprised of a constant function, and a linear function with slope 0.40. If in a given month you use 1000 minutes or fewer, your monthly cost is a constant $50. If you use more than 1000, each additional minute influences the value of C. For example, if you use 1,020 minutes, your cost is: C(1020) = 50+.40(1020-1000) $=$ 50+.4(20) $=$ 50+8 $=$$58.00

It is important to note that in this kind of situation, the time used may to be rounded to the nearest minute. So, for example, if you use 20.5 minutes, you will be charged for 21 minutes. You will be asked to examine this kind of function in one of the review questions.

In the previous lesson, we examined function operations, including composition of functions. We can use composition to model situations that involve more than one function. Consider for example a situation in which you are painting walls. You can express the number of gallons of paint you need for a given wall as a function of the square footage of the wall. If we assume that a gallon of paint can cover approximately 350 square feet, then the equation g(x)= x/350 can be used to model the number of gallons needed for a wall that is x square feet in area. Now you need to buy paint, which usually comes in gallons. If we choose a paint that costs $25 per gallon, then the equation c(g)=25g represents the cost of g gallons of paint. If we compose these functions, we obtain a new function that represents the cost of painting a wall that is x square feet in area: c(g(x))=c (x/350) = 25x/350 = x/14 . When developing a model, it is always a good idea to step back and ask yourself if the specific equation makes sense, and if the type of equation makes sense, given the situation. For example, notice that all three functions involved in the previous example are linear. This makes sense: they all involve a constant rate of change. If we return to the cell phone example, this model makes sense to people who have ever gone over the minutes included in their plan! Finally, if we consider the surface area equation, we are reminded that we have to carefully define the domain of the model. The graph of S(x)= (48/x)+2x2 includes a portion that takes on negative x values, which don’t make sense if we are modeling surface area. Lesson Summary In this lesson we have revisited examples of mathematical models that we have seen throughout this chapter, and we have examined new examples as well. In each example, we considered the details of a given situation, we developed an equation to represent the quantities in the situation, and we analyzed function values to answer questions about the situation. While earlier in the chapter we focused on maximizing and minimizing quantities, here we have also examined other function values. For example, in the example of the ball being tossed into the air, we determined the time at which the ball would hit the ground by calculating the x-intercept of a parabola. In general, using a graphing calculator will help you find function values, including maxima and minima. For the review questions, you are encouraged to use the techniques discussed throughout the chapter, including using the calculator to view a graph, to analyze a table of values, and to identify key values of a function. Points to Consider 1. What aspects of different function families make them appropriate to model different phenomena? 2. How can a graphing calculator help you analyze a model? Review Questions 1. Consider this situation: you run a business making birdhouses. You spend$600 to start your business, and it costs you $5.00 to make each birdhouse. 1. Write a linear equation to represent this situation. 2. State the domain of this function. 3. What does the y-intercept represent? What does the slope represent? 2. Consider this situation: you borrow$500 from a relative, and you agree to pay back the debt at a rate of $15 per month. 1. Write a linear model to represent this situation. 2. Explain why this situation is linear. 3. Graph the function you wrote in part (a) and use the graph to determine the number of months it will take to pay off the debt. 3. You are standing on a wood platform that is 10 feet off the ground. You throw a ball into the air with an initial velocity of 38 ft/s. Write an equation to model the height of the ball as a function of time. How long will it take the ball to reach its maximum height? When will it hit the ground? 4. You are creating a garden in the shape of a triangle. You want the triangle’s height to be 3 feet shorter than the length of the base. You plan to build a fence along the hypotenuse. 1. Write an equation to model the length of the hypotenuse, the longest side of the triangle. 2. What value of x will minimize the length of the hypotenuse? 5. A box is to be made by cutting squares out of the corners of a rectangular piece of cardboard. The dimensions of the cardboard are n inches by m inches. Assume that n>m. 1. Write a model for the volume of the box. 2. What is the largest square that can be cut out of the corners of the cardboard? 6. At your favorite gourmet store, you buy loose tea by the ounce. The store does not calculate for fractions of an ounce. Instead, they round the weight of tea UP to the nearest ounce. Your favorite tea costs$3.00 per ounce. Sketch a function to model the cost of tea as a function of the number of ounces you purchase. Explain why the function is not continuous.
7. The profits for a business can be determined by subtracting the costs from the revenue. Suppose the revenue of a business is modeled by the function R(x) = 5x - 0.01 x2, and the costs of manufacturing the product is modeled by C(x) = 100 + 2x, where x is the number of units of the product.
1. Write a function P(x) to model the company’s profits.
2. Graph P(x) and determine the maximum profit.
8. Express the following situation as a composition of functions: You are running a small business making wooden jewelry boxes. It costs you $5.00 per unit to produce wooden boxes, plus an initial investment of$300 in other materials. It then costs you an additional $2.00 per box to decorate the boxes. 9. Consider the following situation: you are running a business, and you make an initial investment of$1000. It costs you $12 to manufacture each item. 1. Write a function to model your total costs. 2. Write a function to model the average cost per item. Use a graphing utility to determine the minimum of this function. What does this mean? 10. The function f(x) = 0.03x3 - .2x2 + x + 4 is used to model the costs of a running a company that manufacturers book cases. 1. Use a graphing calculator to graph this function. What interval of the domain should you consider? 2. If f represents the costs in thousands, and x represents the number of toys in thousands, what is the cost of manufacturing 2000 toys? How much is this per toy? 3. What is the cost of manufacturing 8000 toys? How much is this per toy? 4. If the cost is$20,000, how many units have been produced?

Feb 23, 2012

Dec 29, 2014