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Chapter 1 | Functions and Graphs
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 Table1.6 , the company arrives at the following quadratic function to model revenue R (in thousands of dollars) 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. a. Predict the revenue if the company sells the item at a price of p =$5 and p =$17. b. Find the zeros of this function and interpret the meaning of the zeros. c. Sketch a graph of R . d. Use the graph to determine the value of p that maximizes revenue. Find the maximum revenue. Solution a. 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 = $141,440. b. 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. c. 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.
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