Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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![**Determining the Rate of Change in Quantity Demanded for a Gaming Console**
The price of a popular gaming console is related to the quantity demanded by the equation:
\[ 2p^2 + x^2 = 41 \]
Here, \( p \) is the price (in hundreds of dollars) and \( x \) is the number of consoles demanded (in thousands).
**Problem Statement:**
Determine how fast the quantity demanded is changing when the price is set at $400 (\( p = 4 \)), the quantity demanded is 3000 (\( x = 3 \)), and the price is decreasing at the rate of $2 per week (\( p' = -0.02 \)).
**Solution:**
First, we differentiate the given equation implicitly with respect to time (\( t \)):
\[ \frac{d}{dt}(2p^2 + x^2) = \frac{d}{dt}(41) \]
This gives us:
\[ 4p \cdot \frac{dp}{dt} + 2x \cdot \frac{dx}{dt} = 0 \]
To find the rate of change of the quantity demanded (\( \frac{dx}{dt} \)), we substitute \( p = 4 \), \( x = 3 \), and \( \frac{dp}{dt} = -0.02 \) into the differentiated equation:
\[ 4(4)(-0.02) + 2(3) \cdot \frac{dx}{dt} = 0 \]
Simplify and solve for \( \frac{dx}{dt} \):
\[ -0.32 + 6 \cdot \frac{dx}{dt} = 0 \]
\[ 6 \cdot \frac{dx}{dt} = 0.32 \]
\[ \frac{dx}{dt} = \frac{0.32}{6} \]
\[ \frac{dx}{dt} = 0.0533 \]
Given that the quantity demanded is in thousands, the rate of change in quantity demanded is approximately 53 units per week. Since \( dx/dt \) is positive, the quantity demanded is increasing.
**Conclusion:**
The quantity demanded is increasing at a rate of approximately 53 units per week.](https://content.bartleby.com/qna-images/question/fbc3f4f4-d793-4827-8e3d-2f51535d2e99/d6e06e68-c17f-4611-a710-bc92dd994b44/xxgeog_processed.jpeg)
Transcribed Image Text:**Determining the Rate of Change in Quantity Demanded for a Gaming Console**
The price of a popular gaming console is related to the quantity demanded by the equation:
\[ 2p^2 + x^2 = 41 \]
Here, \( p \) is the price (in hundreds of dollars) and \( x \) is the number of consoles demanded (in thousands).
**Problem Statement:**
Determine how fast the quantity demanded is changing when the price is set at $400 (\( p = 4 \)), the quantity demanded is 3000 (\( x = 3 \)), and the price is decreasing at the rate of $2 per week (\( p' = -0.02 \)).
**Solution:**
First, we differentiate the given equation implicitly with respect to time (\( t \)):
\[ \frac{d}{dt}(2p^2 + x^2) = \frac{d}{dt}(41) \]
This gives us:
\[ 4p \cdot \frac{dp}{dt} + 2x \cdot \frac{dx}{dt} = 0 \]
To find the rate of change of the quantity demanded (\( \frac{dx}{dt} \)), we substitute \( p = 4 \), \( x = 3 \), and \( \frac{dp}{dt} = -0.02 \) into the differentiated equation:
\[ 4(4)(-0.02) + 2(3) \cdot \frac{dx}{dt} = 0 \]
Simplify and solve for \( \frac{dx}{dt} \):
\[ -0.32 + 6 \cdot \frac{dx}{dt} = 0 \]
\[ 6 \cdot \frac{dx}{dt} = 0.32 \]
\[ \frac{dx}{dt} = \frac{0.32}{6} \]
\[ \frac{dx}{dt} = 0.0533 \]
Given that the quantity demanded is in thousands, the rate of change in quantity demanded is approximately 53 units per week. Since \( dx/dt \) is positive, the quantity demanded is increasing.
**Conclusion:**
The quantity demanded is increasing at a rate of approximately 53 units per week.
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