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7) The total cost for producing x Spikeball sets per day is-x + 8x + 20 dollars, and at this production level each set can be sold for (23 –x) dollars. How many sets should be produced per day to maximize profit? Assume that every set produced is sold. Solution? The profit is (total revenue) – (total production cost). The total revenue is the sale price per set multiplied by the number of sets sold: 23x –x? = 23x –x? - Gx² + 8x + 20) = -x² + 15x – 20 3 x3 15x? |-x² + 15x – 20 dx = 20x + c 2 If we graph this, regardless of the value of c, the local maximum value occurs when x = 18.5, so we say x = 19 as x must be an integer number of Spikeball sets.

7) The total cost for producing x Spikeball sets per day is-x + 8x + 20 dollars, and at this production level each set can be sold for (23 –x) dollars. How many sets should be produced per day to maximize profit? Assume that every set produced is sold. Solution? The profit is (total revenue) – (total production cost). The total revenue is the sale price per set multiplied by the number of sets sold: 23x –x? = 23x –x? - Gx² + 8x + 20) = -x² + 15x – 20 3 x3 15x? |-x² + 15x – 20 dx = 20x + c 2 If we graph this, regardless of the value of c, the local maximum value occurs when x = 18.5, so we say x = 19 as x must be an integer number of Spikeball sets.

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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7) The total cost for producing x Spikeball sets per day is -x² + 8x + 20 dollars, and at this production level each set can be sold for (23 –x) dollars. How many sets should be produced per day to maximize profit? Assume that every set produced is sold. Solution? The profit is (total revenue) – (total production cost). The total revenue is the sale price per set multiplied by the number of sets sold: 23x – x? = 23x-x2-(금x2 + 8x + 20) 3-x2 - Gx² + 8x + 20) = –x² + 15x – 20 х3 15х2 |-x² + 15x – 20 dx = 20x + c 2 If we graph this, regardless of the value of c, the local maximum value occurs when x = 18.5, so we say x = 19 as x must be an integer number of Spikeball sets.
Transcribed Image Text:7) The total cost for producing x Spikeball sets per day is -x² + 8x + 20 dollars, and at this production level each set can be sold for (23 –x) dollars. How many sets should be produced per day to maximize profit? Assume that every set produced is sold. Solution? The profit is (total revenue) – (total production cost). The total revenue is the sale price per set multiplied by the number of sets sold: 23x – x? = 23x-x2-(금x2 + 8x + 20) 3-x2 - Gx² + 8x + 20) = –x² + 15x – 20 х3 15х2 |-x² + 15x – 20 dx = 20x + c 2 If we graph this, regardless of the value of c, the local maximum value occurs when x = 18.5, so we say x = 19 as x must be an integer number of Spikeball sets.
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