(a) Explanation Given information: The monthly demand for a certain brand of perfume is given by the demand equation, p = 100 e − 0.0002 x + 150 (1) Where, p denotes the retail unit price (in dollars). x denotes the quantity (in 1-oz bottles) demanded. Formula used: Chain rule to differentiate the exponential functions is, d d x ( e f ( x ) ) = e f ( x ) ⋅ f ′ ( x ) [ Chain Rule ] The power rule for differentiation of function f ( x ) = x n is, d d x ( x n ) = n x n − 1 The Sum Rule of differentiation is, [ d d x ( f ( x ) ± g ( x ) ) = d d x ( f ( x ) ) ± d d x ( g ( x ) ) ] Calculation: Differentiate both sides of equation (1) with respect to x in to calculate rate of change of the price per bottle, p ′ = d d x ( 100 e − 0.0002 x + 150 ) Use Sum Rule to separate the above derivative, p ′ = d d x ( 100 e − 0.0002 x ) + d d x ( 150 ) Use Chain Rule to simply the differentiation, p ′ = 100 e − 0 (b) To determine The price per bottle when x = 1000 and when x = 2000 .

Applied Calculus for the Managerial, Life, and Social Sciences

10th Edition

ISBN: 9781305887831

Author: Tan

Publisher: CENGAGE CO

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Question

Chapter 5.4, Problem 69E

(a)

To determine

The rate of change of the price per bottle when x=1000 and x=2000 .

(b)

To determine

The price per bottle when x=1000 and when x=2000 .

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Chapter 5.4, Problem 68E

Chapter 5.4, Problem 70E

Chapter 5 Solutions

Applied Calculus for the Managerial, Life, and Social Sciences

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The rate of change of the price per bottle when x = 1000 and x = 2000 .

Solution Summary: The author compares the rate of change of the price per bottle when x=1000 and = 2000.