### Commodity Price Analysis #### Problem Statement: The price of a certain commodity in dollars per unit at time \( t \) (measured in weeks) is given by the equation: \[ p = 18 - 3e^{2t} - 6e^{-t/3} \] **Questions:** (a) What is the price of the commodity at \( t = 0 \)? (b) How fast is the price of the commodity changing at \( t = 0 \)? (c) Find the equilibrium price of the commodity. ### Detailed Analysis: #### (a) Commodity Price at \( t = 0 \): To find the price of the commodity at \( t = 0 \), substitute \( t = 0 \) into the equation \( p = 18 - 3e^{2t} - 6e^{-t/3} \): \[ p = 18 - 3e^{2(0)} - 6e^{-0/3} \] \[ p = 18 - 3e^{0} - 6e^{0} \] \[ p = 18 - 3(1) - 6(1) \] \[ p = 18 - 3 - 6 \] \[ p = 9 \] So, the price of the commodity at \( t = 0 \) is $9. #### (b) Rate of Change of Price at \( t = 0 \): To determine how fast the price of the commodity is changing at \( t = 0 \), we need to find the first derivative of \( p \), denoted \( p' \), and evaluate it at \( t = 0 \). The first derivative of \( p \) with respect to \( t \) is: \[ p' = \frac{d}{dt}(18 - 3e^{2t} - 6e^{-t/3}) \] \[ p' = -3 \cdot \frac{d}{dt}(e^{2t}) - 6 \cdot \frac{d}{dt}(e^{-t/3}) \] Using the chain rule: \[ p' = -3 \cdot 2e^{2t} - 6 \cdot (-\frac{1}{3}e^{-t/3}) \] \[ p' = -6e^{2t} + 2e^{-t/
### Commodity Price Analysis #### Problem Statement: The price of a certain commodity in dollars per unit at time \( t \) (measured in weeks) is given by the equation: \[ p = 18 - 3e^{2t} - 6e^{-t/3} \] **Questions:** (a) What is the price of the commodity at \( t = 0 \)? (b) How fast is the price of the commodity changing at \( t = 0 \)? (c) Find the equilibrium price of the commodity. ### Detailed Analysis: #### (a) Commodity Price at \( t = 0 \): To find the price of the commodity at \( t = 0 \), substitute \( t = 0 \) into the equation \( p = 18 - 3e^{2t} - 6e^{-t/3} \): \[ p = 18 - 3e^{2(0)} - 6e^{-0/3} \] \[ p = 18 - 3e^{0} - 6e^{0} \] \[ p = 18 - 3(1) - 6(1) \] \[ p = 18 - 3 - 6 \] \[ p = 9 \] So, the price of the commodity at \( t = 0 \) is $9. #### (b) Rate of Change of Price at \( t = 0 \): To determine how fast the price of the commodity is changing at \( t = 0 \), we need to find the first derivative of \( p \), denoted \( p' \), and evaluate it at \( t = 0 \). The first derivative of \( p \) with respect to \( t \) is: \[ p' = \frac{d}{dt}(18 - 3e^{2t} - 6e^{-t/3}) \] \[ p' = -3 \cdot \frac{d}{dt}(e^{2t}) - 6 \cdot \frac{d}{dt}(e^{-t/3}) \] Using the chain rule: \[ p' = -3 \cdot 2e^{2t} - 6 \cdot (-\frac{1}{3}e^{-t/3}) \] \[ p' = -6e^{2t} + 2e^{-t/
Calculus: Early Transcendentals
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ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
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![### Commodity Price Analysis
#### Problem Statement:
The price of a certain commodity in dollars per unit at time \( t \) (measured in weeks) is given by the equation:
\[ p = 18 - 3e^{2t} - 6e^{-t/3} \]
**Questions:**
(a) What is the price of the commodity at \( t = 0 \)?
(b) How fast is the price of the commodity changing at \( t = 0 \)?
(c) Find the equilibrium price of the commodity.
### Detailed Analysis:
#### (a) Commodity Price at \( t = 0 \):
To find the price of the commodity at \( t = 0 \), substitute \( t = 0 \) into the equation \( p = 18 - 3e^{2t} - 6e^{-t/3} \):
\[ p = 18 - 3e^{2(0)} - 6e^{-0/3} \]
\[ p = 18 - 3e^{0} - 6e^{0} \]
\[ p = 18 - 3(1) - 6(1) \]
\[ p = 18 - 3 - 6 \]
\[ p = 9 \]
So, the price of the commodity at \( t = 0 \) is $9.
#### (b) Rate of Change of Price at \( t = 0 \):
To determine how fast the price of the commodity is changing at \( t = 0 \), we need to find the first derivative of \( p \), denoted \( p' \), and evaluate it at \( t = 0 \).
The first derivative of \( p \) with respect to \( t \) is:
\[ p' = \frac{d}{dt}(18 - 3e^{2t} - 6e^{-t/3}) \]
\[ p' = -3 \cdot \frac{d}{dt}(e^{2t}) - 6 \cdot \frac{d}{dt}(e^{-t/3}) \]
Using the chain rule:
\[ p' = -3 \cdot 2e^{2t} - 6 \cdot (-\frac{1}{3}e^{-t/3}) \]
\[ p' = -6e^{2t} + 2e^{-t/](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F06486ba7-4300-40ef-aee6-47a68403f9ca%2Fb52d30ce-f5db-4f41-9104-d18c16fa3f41%2Flc1sgzk.jpeg&w=3840&q=75)
Transcribed Image Text:### Commodity Price Analysis
#### Problem Statement:
The price of a certain commodity in dollars per unit at time \( t \) (measured in weeks) is given by the equation:
\[ p = 18 - 3e^{2t} - 6e^{-t/3} \]
**Questions:**
(a) What is the price of the commodity at \( t = 0 \)?
(b) How fast is the price of the commodity changing at \( t = 0 \)?
(c) Find the equilibrium price of the commodity.
### Detailed Analysis:
#### (a) Commodity Price at \( t = 0 \):
To find the price of the commodity at \( t = 0 \), substitute \( t = 0 \) into the equation \( p = 18 - 3e^{2t} - 6e^{-t/3} \):
\[ p = 18 - 3e^{2(0)} - 6e^{-0/3} \]
\[ p = 18 - 3e^{0} - 6e^{0} \]
\[ p = 18 - 3(1) - 6(1) \]
\[ p = 18 - 3 - 6 \]
\[ p = 9 \]
So, the price of the commodity at \( t = 0 \) is $9.
#### (b) Rate of Change of Price at \( t = 0 \):
To determine how fast the price of the commodity is changing at \( t = 0 \), we need to find the first derivative of \( p \), denoted \( p' \), and evaluate it at \( t = 0 \).
The first derivative of \( p \) with respect to \( t \) is:
\[ p' = \frac{d}{dt}(18 - 3e^{2t} - 6e^{-t/3}) \]
\[ p' = -3 \cdot \frac{d}{dt}(e^{2t}) - 6 \cdot \frac{d}{dt}(e^{-t/3}) \]
Using the chain rule:
\[ p' = -3 \cdot 2e^{2t} - 6 \cdot (-\frac{1}{3}e^{-t/3}) \]
\[ p' = -6e^{2t} + 2e^{-t/
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