Lecture 5.6

# Lecture 5.6 - Chapter 5 Graphing and Optimization Section 6...

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Chapter 5 Graphing and Optimization Section 6 Optimization

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2 Barnett/Ziegler/Byleen Business Calculus 12e Objectives for Section 5.6 Optimization The student will be able to calculate: Area and perimeter Revenue and profit Inventory control.
3 Area Problems 2 2 or 2 A=xy P xy Px y = + = + 22 a bc = 2 2 24 SA x xy V xy = = x x y

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4 Barnett/Ziegler/Byleen Business Calculus 12e The techniques used to solve optimization problems are best illustrated through examples. Let’s begin with a geometric example. Find the dimensions of a rectangular area of 225 square meters that has the least perimeter. Hint: Drawing a figure might help. Area and Perimeter Example 1
5 Barnett/Ziegler/Byleen Business Calculus 12e Let L = length, W = width. The formulas for area and perimeter are A = L · W = 225 P = 2 · L + 2 · W From the area equation solve for L and substitute that value of L into the perimeter equation to get an equation in one unknown: Area and Perimeter Example 1 L W W W P W L 2 450 225 + = =

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6 Barnett/Ziegler/Byleen Business Calculus 12e Example 1 (continued) We wish to minimize P , so we take the derivative and look at the critical values. There is a critical value at 15. (Disregard W = –15 since the width cannot be negative.) 2 2 2 ) 15 ( ) 15 ( 2 450 2 ) ( ' W W W W W W P + = =
7 Barnett/Ziegler/Byleen Business Calculus 12e Example 1 (continued) P ′′ (15) > 0, so this is a local minimum. The least perimeter occurs when W = 15.

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Lecture 5.6 - Chapter 5 Graphing and Optimization Section 6...

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