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Solve from 9th question explaining every single calculation 

Let’s solve the questions starting from Question 9. I'll break down each step of the calculations and provide clear explanations.

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### **Question 9**:  
The cost function for \( x \) units is given by:  
\[
C(x) = ax^3 + bx^2 - cx + d, \quad \text{where } a > 0, b < 0, c > 0
\]  
We are asked to show that **average variable cost (AVC)** and **marginal cost (MC)** intersect at the minimum of the **average variable cost**.

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#### **Step 1: Definitions**  

1. **Total Cost (TC):** Given as \( C(x) = ax^3 + bx^2 - cx + d \).  
   Here, \( d \) is the fixed cost (constant term).

2. **Variable Cost (VC):** Variable cost is the part of total cost dependent on \( x \), so:  
   \[
   VC(x) = ax^3 + bx^2 - cx
   \]

3. **Average Variable Cost (AVC):**  
   Average Variable Cost is:  
   \[
   AVC(x) = \frac{VC(x)}{x} = \frac{ax^3 + bx^2 - cx}{x} = ax^2 + bx - c
   \]

4. **Marginal Cost (MC):**  
   Marginal Cost is the derivative of total cost \( C(x) \) with respect to \( x \):  
   \[
   MC(x) = \frac{d}{dx} \big[ax^3 + bx^2 - cx + d\big] = 3ax^2 + 2bx - c
   \]

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#### **Step 2: Finding Intersection Point**

To find the intersection of \( AVC(x) \) and \( MC(x) \), set \( AVC(x) = MC(x) \):  
\[
ax^2 + bx - c = 3ax^2 + 2bx - c
\]

Simplify by subtracting \( ax^2 + bx - c \) from both sides:  
\[
0 = 2ax^2 + bx
\]

Factorize:  
\[
x(2ax + b) = 0
\]

Thus, \( x = 0 \) or \( x = -\frac{b}{2a} \).  
Since \( x = 0 \) is not feasible for production, the intersection occurs at:  
\[
x = -\frac{b}{2a}
\]

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#### **Step 3: Verify Minimum Point**

To verify that this is the minimum of \( AVC(x) \), compute the derivative of \( AVC(x) \):  
\[
AVC(x) = ax^2 + bx - c
\]
\[
\frac{d}{dx} [AVC(x)] = 2ax + b
\]

Set \( \frac{d}{dx} [AVC(x)] = 0 \) for critical points:  
\[
2ax + b = 0 \implies x = -\frac{b}{2a}
\]

Thus, \( AVC(x) \) is minimum at \( x = -\frac{b}{2a} \), which is the same point where \( AVC(x) = MC(x) \).

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### **Conclusion for Question 9**:  
The average variable cost (AVC) and marginal cost (MC) curves intersect at the minimum of the average variable cost curve.

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Would you like to proceed with Question 10 or review any steps in this solution? Let me know!  

Here are **five related questions** you might consider:  
1. What is the significance of the intersection of AVC and MC?  
2. How does fixed cost impact the cost functions?  
3. Why does \( AVC(x) \) reach a minimum?  
4. How does the shape of \( AVC(x) \) change for different values of \( a, b, c \)?  
5. Can AVC and MC fail to intersect for certain cost functions?  

**Tip:** Always verify minimum/maximum points by checking the derivative's sign or the second derivative.

Solve from 9th question explaining every single calculation.

Explore a step-by-step solution to show the intersection of average variable cost (AVC) and marginal cost (MC) at the minimum point of AVC. Learn key calculus concepts, including derivatives and optimization, suitable for Grades 11-12 or introductory college-level calculus students.

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