Amit sells pakodas in boxes of different sizes containing different numbers of pakodas. The pakodas are priced at Rs. 3 per pakoda up to 300 pakodas. For every additional 25 pakodas, the price of the whole lot goes down by 10 paise per pakoda. What should be the number of pakodas in the box that would maximize his revenue

To determine the number of pakodas that Amit should include in a box to maximize his revenue, we need to analyze how the price per pakoda decreases as the number of pakodas increases.

### Key Details
- Up to 300 pakodas: Price per pakoda = Rs. 3
- For every additional 25 pakodas beyond 300, the price of the entire lot decreases by Rs. 0.10 (10 paise) per pakoda.

Let:
- \( x \) represent the number of pakodas sold in the box.
- \( p(x) \) represent the price per pakoda.

### Setup
1. **Price Function**:  
   For the first 300 pakodas, the price per pakoda remains Rs. 3. After 300 pakodas, for every 25 pakodas, the price per pakoda decreases by 10 paise. Thus, the price function for \( x \geq 300 \) can be written as:
   \[
   p(x) = 3 - 0.10 \times \frac{x - 300}{25}
   \]
   Simplifying:
   \[
   p(x) = 3 - 0.004(x - 300)
   \]
   This gives us the price per pakoda for \( x \geq 300 \).

2. **Revenue Function**:  
   Revenue, \( R(x) \), is calculated as:
   \[
   R(x) = x \cdot p(x)
   \]
   Substituting the price function:
   \[
   R(x) = x \left(3 - 0.004(x - 300)\right)
   \]
   Expand and simplify:
   \[
   R(x) = x \cdot \left(3 - 0.004x + 1.2\right)
   \]
   \[
   R(x) = x \cdot \left(4.2 - 0.004x\right)
   \]
   \[
   R(x) = 4.2x - 0.004x^2
   \]

3. **Maximization**:  
   To find the number of pakodas that maximizes the revenue, we need to differentiate \( R(x) \) with respect to \( x \), set it equal to zero, and solve for \( x \).

   The first derivative of the revenue function is:
   \[
   \frac{dR}{dx} = 4.2 - 0.008x
   \]
   Set \( \frac{dR}{dx} = 0 \):
   \[
   4.2 - 0.008x = 0
   \]
   Solving for \( x \):
   \[
   0.008x = 4.2
   \]
   \[
   x = \frac{4.2}{0.008} = 525
   \]

### Conclusion
Amit should include **525 pakodas** in the box to maximize his revenue.

---

Would you like further clarification or details? 

Here are five related questions to expand your understanding:
1. How would the total revenue change if Amit sells only 300 pakodas?
2. What happens to the revenue if the price drop increases to 20 paise per 25 pakodas?
3. How does the revenue function behave for more than 600 pakodas?
4. Can you derive the break-even point where Amit's revenue stops increasing?
5. How would adding a fixed cost per box affect the optimal number of pakodas?

**Tip**: Revenue maximization often involves balancing volume sold with price per unit; understanding this trade-off is crucial in real-world pricing decisions.

Discover how to maximize revenue by optimizing the number of pakodas per box. Learn about the price function, revenue function, and derivative analysis to find the optimal number of pakodas. Suitable for advanced high school students and anyone interested in revenue optimization through mathematics.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation