SECTION 3.4

Build Quadratic Models from Verbal Descriptions and from Data

159

3.4 Build Quadratic Models from Verbal Descriptions and from Data
PREPARING FOR THIS SECTION Before getting started, review the following:
• Problem Solving (Appendix A, Section A.8, pp. A63 –A70)

• Linear Models: Building Linear Functions from
Data (Section 3.2, pp. 140–143)

Now Work the ‘Are You Prepared?’ problems on page 164.

OBJECTIVES 1 Build Quadratic Models from Verbal Descriptions (p. 159)
2 Build Quadratic Models from Data (p. 163)

In this section we will first discuss models in the form of a quadratic function when a verbal description of the problem is given. We end the section by fitting a quadratic function to data, which is another form of modeling.
When a mathematical model is in the form of a quadratic function, the properties of the graph of the quadratic function can provide important information about the model. In particular, we can use the quadratic function to determine the maximum or minimum value of the function. The fact that the graph of a quadratic function has a maximum or minimum value enables us to answer questions involving optimization, that is, finding the maximum or minimum values in models.

1 Build Quadratic Models from Verbal Descriptions
In economics, revenue R, in dollars, is defined as the amount of money received from the sale of an item and is equal to the unit selling price p, in dollars, of the item times the number x of units actually sold. That is,
R = xp
The Law of Demand states that p and x are related: As one increases, the other decreases. The equation that relates p and x is called the demand equation. When the demand equation is linear, the revenue model is a quadratic function.

EXAM PL E 1

Maximizing Revenue
The marketing department at Texas Instruments has found that, when certain calculators are sold at a price of p dollars per unit, the number x of calculators sold is given by the demand equation x = 21,000 - 150p
(a)
(b)
(c)
(d)
(e)
(f)
(g)

Solution

Find a model that expresses the revenue R as a function of the price p.
What is the domain of R?
What unit price should be used to maximize revenue?
If this price is charged, what is the maximum revenue?
How many units are sold at this price?
Graph R.
What price should Texas Instruments charge to collect at least $675,000 in revenue? (a) The revenue R is R = xp, where x = 21,000 - 150p.
R = xp = (21,000 - 150p)p = -150p 2 + 21,000p

The Model

(b) Because x represents the number of calculators sold, we have x Ú 0, so
21,000 - 150p Ú 0. Solving this linear inequality, we find that p … 140. In addition, Texas Instruments will only charge a positive price for the calculator, so p 7 0. Combining these inequalities, the domain of R is 5 p ͉ 0 6 p … 140 6 .

Linear and Quadratic Functions

(c) The function R is a quadratic function with a = -150, b = 21,000, and c = 0.
Because a 6 0, the vertex is the highest point on the parabola. The revenue R is a maximum when the price p is p = -

21,000
21,000
b
= = = +70.00
2a c 2( -150)
-300

a = -150, b = 21,000

(d) The maximum revenue R is
R(70) = -150(70)2 + 21,000(70) = +735,000
(e) The number of calculators sold is given by the demand equation x = 21,000 - 150p. At a price of p = +70, x = 21,000 - 150(70) = 10,500 calculators are sold.
(f) To graph R, plot the intercept (140, 0) and the vertex (70, 735,000). See Figure 24 for the graph.
Figure 24
Revenue (dollars)

CHAPTER 3

R
800,000
700,000
600,000
500,000
400,000
300,000
200,000
100,000

(70, 735,000)

0

14 28 42 56 70 84 98 112 126 140
Price p per calculator (dollars)

p

(g) Graph R = 675,000 and R 1p2 = -150p 2 + 21,000p on the same Cartesian plane. See Figure 25. We find where the graphs intersect by solving
675,000
2
150p - 21,000p + 675,000 p 2 - 140p + 4500
1p - 502 1p - 902 p = 50 or p
Figure 25