TAILIEUCHUNG - Calculus and its applications: 2.5

"Calculus and its applications: " - Maximum-Minimum Problems-Business and Economics applications have objective: solve maximum and minimum problems using calculus. | 2012 Pearson Education, Inc. All rights reserved Slide Maximum-Minimum Problems; Business and Economics Applications OBJECTIVE Solve maximum and minimum problems using calculus. 2012 Pearson Education, Inc. All rights reserved Slide A Strategy for Solving Maximum-Minimum Problems: 1. Read the problem carefully. If relevant, make a drawing. 2. Make a list of appropriate variables and constants, noting what varies, what stays fixed, and what units are used. Label the measurements on your drawing, if one exists. Maximum-Minimum Problems; Business and Economics Applications 2012 Pearson Education, Inc. All rights reserved Slide A Strategy for Solving Maximum-Minimum Problems (concluded): 3. Translate the problem to an equation involving a quantity Q to be maximized or minimized. Try to represent Q in terms of the variables of step (2). to express Q as a function of one variable. Use the procedures developed in sections – to determine the maximum or minimum values and the points at which they occur. Maximum-Minimum Problems; Business and Economics Applications 2012 Pearson Education, Inc. All rights reserved Slide Example 1: From a thin piece of cardboard 8 in. by 8 in., square corners are cut out so that the sides can be folded up to make a box. What dimensions will yield a box of maximum volume? What is the maximum volume? 1st make a drawing in which x is the length of each square to be cut. Maximum-Minimum Problems; Business and Economics Applications 2012 Pearson Education, Inc. All rights reserved Slide Example 1 (continued): 2nd write an equation for the volume of the box. Note that x must be between 0 and 4. So, we need to maximize the volume equation on the interval (0, 4). Maximum-Minimum Problems; Business and Economics Applications 2012 Pearson Education, Inc. All rights reserved Slide Example 1 (continued): is the only critical value in (0, 4). So, we can use the second .

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