An Appliance Store Knows That If It Sells Its New Gadgets For R50 Each, It Can Sell 200 Per Month, And If It Sells The Same Gadgets For R40, It Will Sell 300 Per Month. Let $p$ Represent The Price Of The Gadgets And Let $n$

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Introduction

In the world of business, pricing is a crucial aspect that can make or break a product's success. An appliance store, for instance, needs to find the perfect balance between the price of its gadgets and the number of units it sells. In this article, we will delve into a mathematical problem that illustrates the relationship between price and quantity sold, and explore how to find the optimal price for the store's gadgets.

The Problem

An appliance store knows that if it sells its new gadgets for R50 each, it can sell 200 per month, and if it sells the same gadgets for R40, it will sell 300 per month. Let $p$ represent the price of the gadgets and let $n$ represent the number of units sold. We can represent the two scenarios as follows:

  • If the price is R50, the number of units sold is 200, so we can write the equation: $50n = 200$
  • If the price is R40, the number of units sold is 300, so we can write the equation: $40n = 300$

Solving the Equations

To find the relationship between the price and the number of units sold, we can solve the two equations simultaneously. We can start by dividing both sides of the first equation by 50 to isolate $n$:

n=20050n = \frac{200}{50}

n=4n = 4

This tells us that if the price is R50, the store sells 4 units per month. Now, we can substitute this value of $n$ into the second equation to find the price:

40n=30040n = 300

40(4)=30040(4) = 300

160=300160 = 300

This equation is not true, which means that the store cannot sell 300 units per month at a price of R40. However, we can use this information to find the relationship between the price and the number of units sold.

Finding the Relationship

Let's go back to the two equations:

  • 50n=20050n = 200

  • 40n=30040n = 300

We can divide both sides of the first equation by 50 to get:

n=20050n = \frac{200}{50}

n=4n = 4

This tells us that if the price is R50, the store sells 4 units per month. Now, we can divide both sides of the second equation by 40 to get:

n=30040n = \frac{300}{40}

n=7.5n = 7.5

This tells us that if the price is R40, the store sells 7.5 units per month. However, we know that the store cannot sell a fraction of a unit, so we can round down to 7 units per month.

Graphing the Relationship

To visualize the relationship between the price and the number of units sold, we can graph the two equations on a coordinate plane. The x-axis represents the price, and the y-axis represents the number of units sold.

  • The first equation is $50n = 200$, which can be rewritten as $n = \frac{200}{50}$
  • The second equation is $40n = 300$, which can be rewritten as $n = \frac{300}{40}$

We can plot the two points on the graph:

  • (50, 4)
  • (40, 7.5)

We can then draw a line through the two points to represent the relationship between the price and the number of units sold.

Finding the Optimal Price

To find the optimal price, we need to find the point on the graph where the number of units sold is maximized. We can do this by finding the vertex of the parabola represented by the graph.

The vertex of a parabola is the point where the parabola changes direction, and it is represented by the equation:

x=−b2ax = -\frac{b}{2a}

In this case, the equation is:

n=20050n = \frac{200}{50}

n=4n = 4

This tells us that if the price is R50, the store sells 4 units per month. However, we know that the store cannot sell a fraction of a unit, so we can round down to 3 units per month.

Conclusion

In conclusion, we have analyzed the relationship between the price and the number of units sold for an appliance store's gadgets. We have found that if the price is R50, the store sells 4 units per month, and if the price is R40, the store sells 7.5 units per month. We have also graphed the relationship between the price and the number of units sold, and we have found the optimal price to be R50.

Optimal Price Calculation

To calculate the optimal price, we can use the following formula:

p=200np = \frac{200}{n}

We can substitute the value of $n$ into the formula to get:

p=2004p = \frac{200}{4}

p=50p = 50

This tells us that the optimal price is R50.

Real-World Applications

The concept of finding the optimal price is not limited to the appliance store's problem. It can be applied to various real-world scenarios, such as:

  • Pricing strategy for a product
  • Determining the optimal price for a service
  • Finding the best price for a commodity

Limitations

The analysis presented in this article has some limitations. For instance:

  • The problem assumes that the store sells the same number of units per month, which may not be the case in reality.
  • The problem assumes that the store sells the same type of gadgets, which may not be the case in reality.
  • The problem assumes that the store has a fixed cost, which may not be the case in reality.

Future Research Directions

Future research directions for this problem could include:

  • Developing a more realistic model that takes into account the store's fixed costs and variable costs.
  • Investigating the impact of different pricing strategies on the store's revenue and profit.
  • Analyzing the relationship between the price and the number of units sold for different types of products.

Conclusion

In conclusion, we have analyzed the relationship between the price and the number of units sold for an appliance store's gadgets. We have found that if the price is R50, the store sells 4 units per month, and if the price is R40, the store sells 7.5 units per month. We have also graphed the relationship between the price and the number of units sold, and we have found the optimal price to be R50. The concept of finding the optimal price is not limited to the appliance store's problem, and it can be applied to various real-world scenarios.

Introduction

In our previous article, we analyzed the relationship between the price and the number of units sold for an appliance store's gadgets. We found that if the price is R50, the store sells 4 units per month, and if the price is R40, the store sells 7.5 units per month. We also graphed the relationship between the price and the number of units sold, and we found the optimal price to be R50. In this article, we will answer some frequently asked questions related to the appliance store's pricing puzzle.

Q: What is the optimal price for the appliance store's gadgets?

A: The optimal price for the appliance store's gadgets is R50.

Q: Why is R50 the optimal price?

A: R50 is the optimal price because it maximizes the number of units sold. If the price is R50, the store sells 4 units per month, which is the highest number of units sold among all possible prices.

Q: What happens if the price is higher than R50?

A: If the price is higher than R50, the store will sell fewer units per month. For example, if the price is R60, the store will sell 3 units per month.

Q: What happens if the price is lower than R40?

A: If the price is lower than R40, the store will sell more units per month. For example, if the price is R30, the store will sell 10 units per month.

Q: Can the store sell a fraction of a unit?

A: No, the store cannot sell a fraction of a unit. If the price is R40, the store will sell 7 units per month, not 7.5 units per month.

Q: How does the store's fixed cost affect the optimal price?

A: The store's fixed cost does not affect the optimal price. The optimal price is determined by the relationship between the price and the number of units sold, not by the store's fixed cost.

Q: Can the store change its pricing strategy to increase revenue?

A: Yes, the store can change its pricing strategy to increase revenue. For example, the store can offer discounts to customers who buy multiple units, or it can offer premium pricing for high-end products.

Q: How does the store's variable cost affect the optimal price?

A: The store's variable cost does not affect the optimal price. The optimal price is determined by the relationship between the price and the number of units sold, not by the store's variable cost.

Q: Can the store sell different types of products at different prices?

A: Yes, the store can sell different types of products at different prices. For example, the store can sell high-end products at a premium price and low-end products at a discount price.

Q: How does the store's pricing strategy affect its profit?

A: The store's pricing strategy affects its profit by determining the number of units sold and the revenue generated. If the store sets a high price, it may sell fewer units, but it may also generate more revenue per unit.

Q: Can the store use data analytics to inform its pricing strategy?

A: Yes, the store can use data analytics to inform its pricing strategy. For example, the store can use data on customer behavior and market trends to determine the optimal price for its products.

Conclusion

In conclusion, we have answered some frequently asked questions related to the appliance store's pricing puzzle. We have found that the optimal price for the store's gadgets is R50, and we have discussed how the store's pricing strategy can affect its revenue and profit. We have also discussed how the store can use data analytics to inform its pricing strategy and increase revenue.