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1.2 Basic classes of functions  (Page 5/28)

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Consider the quadratic function f ( x ) = 3 x 2 − 6 x + 2 . Find the zeros of f . Does the parabola open upward or downward?

The zeros are x = 1 ± 3 / 3 . The parabola opens upward.

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Mathematical models

A large variety of real-world situations can be described using mathematical models . A mathematical model is a method of simulating real-life situations with mathematical equations. Physicists, engineers, economists, and other researchers develop models by combining observation with quantitative data to develop equations, functions, graphs, and other mathematical tools to describe the behavior of various systems accurately. Models are useful because they help predict future outcomes. Examples of mathematical models include the study of population dynamics, investigations of weather patterns, and predictions of product sales.

As an example, let’s consider a mathematical model that a company could use to describe its revenue for the sale of a particular item. The amount of revenue R a company receives for the sale of n items sold at a price of p dollars per item is described by the equation R = p · n . The company is interested in how the sales change as the price of the item changes. Suppose the data in [link] show the number of units a company sells as a function of the price per item.

Number of units sold n (in thousands) as a function of price per unit p (in dollars)
p 6 8 10 12 14
n 19.4 18.5 16.2 13.8 12.2

In [link] , we see the graph the number of units sold (in thousands) as a function of price (in dollars). We note from the shape of the graph that the number of units sold is likely a linear function of price per item, and the data can be closely approximated by the linear function n = −1.04 p + 26 for 0 ≤ p ≤ 25 , where n predicts the number of units sold in thousands. Using this linear function, the revenue (in thousands of dollars) can be estimated by the quadratic function

R ( p ) = p · ( −1.04 p + 26 ) = −1.04 p 2 + 26 p

for 0 ≤ p ≤ 25 . In [link] , we use this quadratic function to predict the amount of revenue the company receives depending on the price the company charges per item. Note that we cannot conclude definitively the actual number of units sold for values of p , for which no data are collected. However, given the other data values and the graph shown, it seems reasonable that the number of units sold (in thousands) if the price charged is p dollars may be close to the values predicted by the linear function n = −1.04 p + 26 .

An image of a graph. The y axis runs from 0 to 28 and is labeled “n, units sold in thousands”. The x axis runs from 0 to 28 and is labeled “p, price in dollars”. The graph is of the function “n = -1.04p + 26”, which is a decreasing line function that starts at the y intercept point (0, 26). There are 5 points plotted on the graph at (6, 19.4), (8, 18.5), (10, 16.2), (12, 13.8), and (14, 12.2). The points are not on the graph of the function line, but are very close to it. The function has an x intercept at the point (25, 0).
The data collected for the number of items sold as a function of price is roughly linear. We use the linear function n = −1.04 p + 26 to estimate this function.

Maximizing revenue

A company is interested in predicting the amount of revenue it will receive depending on the price it charges for a particular item. Using the data from [link] , the company arrives at the following quadratic function to model revenue R as a function of price per item p :

R ( p ) = p · ( −1.04 p + 26 ) = −1.04 p 2 + 26 p

for 0 ≤ p ≤ 25 .

  1. Predict the revenue if the company sells the item at a price of p = $ 5 and p = $ 17 .
  2. Find the zeros of this function and interpret the meaning of the zeros.
  3. Sketch a graph of R .
  4. Use the graph to determine the value of p that maximizes revenue. Find the maximum revenue.
  1. Evaluating the revenue function at p = 5 and p = 17 , we can conclude that
    R ( 5 ) = −1.04 ( 5 ) 2 + 26 ( 5 ) = 104 , so revenue = $104,000; R ( 17 ) = −1.04 ( 17 ) 2 + 26 ( 17 ) = 141.44 , so revenue = $144,440.
  2. The zeros of this function can be found by solving the equation −1.04 p 2 + 26 p = 0 . When we factor the quadratic expression, we get p ( −1.04 p + 26 ) = 0 . The solutions to this equation are given by p = 0 , 25 . For these values of p , the revenue is zero. When p = $ 0 , the revenue is zero because the company is giving away its merchandise for free. When p = $ 25 , the revenue is zero because the price is too high, and no one will buy any items.
  3. Knowing the fact that the function is quadratic, we also know the graph is a parabola. Since the leading coefficient is negative, the parabola opens downward. One property of parabolas is that they are symmetric about the axis, so since the zeros are at p = 0 and p = 25 , the parabola must be symmetric about the line halfway between them, or p = 12.5 .
    An image of a graph. The y axis runs from 0 to 170 and is labeled “R, revenue in thousands of dollars”. The x axis runs from 0 to 28 and is labeled “p, price in dollars”. The graph is of the function “n = -1.04(p squared) + 26p”, which is a parabola that starts at the origin. The function increases until the maximum point at (12.5, 162.5) and then begins decreasing. The function has x intercepts at the origin and the point (25, 0). The y intercept is at the origin.
  4. The function is a parabola with zeros at p = 0 and p = 25 , and it is symmetric about the line p = 12.5 , so the maximum revenue occurs at a price of p = $ 12.50 per item. At that price, the revenue is R ( p ) = −1.04 ( 12.5 ) 2 + 26 ( 12.5 ) = $ 162 , 500 .
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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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