The price of a popular gaming console is related to the quantity demanded by the equation 2p² + x² = 41 where p is the price (in hundreds of dollars) and x is the number of consoles demanded (in thousands). 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). Give your answer as a whole number (remember that the quantity demanded is measured in thousands). Make sure to indicate whether it is increasing or decreasing and include the correct units.
The price of a popular gaming console is related to the quantity demanded by the equation 2p² + x² = 41 where p is the price (in hundreds of dollars) and x is the number of consoles demanded (in thousands). 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). Give your answer as a whole number (remember that the quantity demanded is measured in thousands). Make sure to indicate whether it is increasing or decreasing and include the correct units.
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffbc3f4f4-d793-4827-8e3d-2f51535d2e99%2Fd6e06e68-c17f-4611-a710-bc92dd994b44%2Fxxgeog_processed.jpeg&w=3840&q=75)
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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