Chapter 4, Problem 84RE

### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

Chapter
Section

### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# Demand The demand function for a product is given by p = 9000 ( 1 − 4 4 + e − 0.003 x ) where p is the price per unit (in dollars) and x is the number of units sold. Find the number of units sold when (a) p = $200 and (b) p =$1000.

(a)

To determine

To calculate: The number of units sold if the price of the product is $200, when the demand function for a product is, p=9000(144+e0.003x). Explanation Given Information: The demand function for a product is, p=9000(144+e0.003x) Where, p is the price of the product in dollars and x is the number of units sold. Formula used: The inverse property of logarithms for the expression lnex is lnex=x. The property of logarithms for the expression ln(xy) is ln(xy)=lnxlny where x and y are real numbers greater than 0. Calculation: Consider the demand function, p=9000(144+e0.003x). Calculate the number of units sold if the price of the product is$200 by substituting p=200 in the demand function as,

200=9000(144+e0.003x)

Divide the both sides by 9000 as,

2009000=9000(144+e0.003x)9000144+e0.003x=290

Subtract 1 from both sides as,

144+e0.003x1=290144+e0.003x=2909044+e0.003x=8890

Cancel the minus sign from the both sides and simplify as,

44+e0.003x=4445

Multiply both sides by 45(4+e0

(b)

To determine

To calculate: The number of units sold if the price of the product is \$1000, when the demand function for a product is p=9000(144+e0.003x).

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