For the following demand​ function, find a.​ E, and b. the values of q​ (if any) at which total revenue is maximized.
q equals 37 comma 400 minus 3 p squared
Question content area bottom
Part 1
a. Determine the elasticity of​ demand, E. 
Eequals
  
StartFraction 6 p squared Over 37400 minus 3 p squared EndFraction
 ​(Type an expression using p as the​ variable.) 
Part 2
b. Determine the value of q that maximizes the revenue. Select the correct choice​ below, and if​ necessary, fill in the answer box within your choice.
A.
Total revenue is maximized at about qequals
  
enter your response here
​(Round to the nearest whole number as​ needed.) 
B.
No values of q maximize total revenue. 

Let's work through the problem step-by-step.

---

### Part 1: Elasticity of Demand (E)
The **demand function** is given by:  
\[
q = 37,400 - 3p^2
\]  
Elasticity of demand, \(E\), is generally defined as:
\[
E = \frac{p}{q} \cdot \frac{dq}{dp}
\]

1. **Find the derivative** of \(q\) with respect to \(p\):
   \[
   \frac{dq}{dp} = -6p
   \]

2. **Substitute** the values of \(q\) and \(\frac{dq}{dp}\) into the formula for \(E\):
   \[
   E = \frac{p}{37,400 - 3p^2} \cdot (-6p) = \frac{-6p^2}{37,400 - 3p^2}
   \]

Thus, the elasticity of demand is:
\[
E = \frac{6p^2}{37,400 - 3p^2}
\]

---

### Part 2: Values of \(q\) that Maximize Total Revenue  
**Total revenue** is given by:
\[
TR = p \cdot q
\]
Since \(q = 37,400 - 3p^2\), we substitute this into the expression for total revenue:
\[
TR = p \cdot (37,400 - 3p^2) = 37,400p - 3p^3
\]

Now, to find the value of \(p\) that maximizes total revenue, we take the derivative of \(TR\) with respect to \(p\) and set it equal to 0.

1. **Differentiate** \(TR\) with respect to \(p\):
   \[
   \frac{d(TR)}{dp} = 37,400 - 9p^2
   \]

2. **Set the derivative equal to 0**:
   \[
   37,400 - 9p^2 = 0
   \]
   \[
   9p^2 = 37,400
   \]
   \[
   p^2 = \frac{37,400}{9} = 4,155.56
   \]
   \[
   p = \sqrt{4,155.56} \approx 64.5
   \]

3. **Find the corresponding value of \(q\)**:
   Substituting \(p = 64.5\) into the demand function:
   \[
   q = 37,400 - 3(64.5)^2 = 37,400 - 12,495.75 = 24,904.25 \approx 24,904
   \]

So, the value of \(q\) that maximizes total revenue is approximately:
\[
q \approx 24,904
\]

---

### Final Answers:
**Part 1:**  
\[
E = \frac{6p^2}{37,400 - 3p^2}
\]

**Part 2:**  
A. Total revenue is maximized at approximately \(q = 24,904\). 

---

### Follow-up Questions:
1. How does elasticity of demand relate to maximizing revenue?
2. What happens to total revenue if \(p\) is increased beyond 64.5?
3. Can we use the second derivative test to confirm that this value of \(p\) gives a maximum?
4. How would the result change if the demand function had a linear form?
5. How does elasticity change as the price \(p\) increases?

---

### Tip:
The total revenue is maximized when the price elasticity of demand is equal to 1 (unitary elastic).

For the following demand function, find a.​ E, and b. the values of q​ (if any) at which total revenue is maximized.
q = 37,400 - 3p²

Part 1
a. Determine the elasticity of​ demand, E.
E = (6p²) / (37,400 - 3p²)

Part 2
b. Determine the value of q that maximizes the revenue.
A. Total revenue is maximized at about q = [answer].
B. No values of q maximize total revenue.

Explore the step-by-step solution to finding the elasticity of demand and revenue maximization for the function q = 37,400 - 3p². Learn to apply calculus concepts, including derivatives, and understand how optimization techniques identify maximum revenue points. Suitable for college-level economics or calculus courses.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation