   Chapter 4.4, Problem 85E ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

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 = 6000 ( 1 − 4 4 + e − 0.002 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 =$700. Use a graphing utility to verify your results in parts (a) and (b).

(a)

To determine

To calculate: The number of units sold if the price of the product is $200 and the demand function is given by: p=6000(144+e0.002x) Where, p is the price of the product in dollars and x is the number of units sold. Also verify the result using technology. Explanation Given Information: The demand function for a product is given by: p=6000(144+e0.002x) 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=6000(144+e0.002x) Calculate the number of units sold if the price of the product is$200 by substituting p=200 in the demand function as,

200=6000(144+e0.002x)

Divide the both sides by 6000 as,

2006000=6000(144+e0.002x)6000144+e0.002x=130

Subtract 1 from both sides as,

144+e0.002x1=130144+e0.002x=1303044+e0.002x=2930

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

44+e0.002x=2930

Multiply both sides by 30(4+e0.002x) as,

44+e0.002x[30(4+e0

(b)

To determine

To calculate: The number of units sold if the price of the product is $700, that is p=$700 when the demand function for a product is given by:

p=6000(144+e0.002x)

Where, p is the price of the product in dollars and x is the number of units sold. Also verify the result using technology.

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