Cost Of Digging A Well Increasing After Every Meter A Mathematical Analysis

In this article, we delve into a fascinating mathematical problem concerning the cost of digging a well. The cost structure presents an interesting scenario where the price escalates with depth, creating an arithmetic progression. We will explore the intricacies of this cost model, derive formulas for calculating the cost at any given depth, and discuss various applications of this concept. This exploration will not only enhance our understanding of arithmetic progressions but also provide practical insights into real-world cost calculations. Understanding the cost of digging a well as it increases with depth involves applying concepts of arithmetic progression. Arithmetic progressions are sequences where the difference between consecutive terms remains constant. In this specific scenario, the initial cost for the first meter is ₹150, and the cost increases by ₹50 for each subsequent meter. This consistent increase defines the common difference in our arithmetic progression, which is crucial for calculating the overall cost at various depths. The problem allows us to apply these mathematical principles to a tangible, real-world situation. This helps in understanding not only the theoretical aspects of arithmetic progressions but also their practical applications in cost estimation and financial planning. This combination of theoretical knowledge and practical application is key to mastering mathematical concepts and their relevance in everyday life. Moreover, this analysis will illustrate how mathematical models can accurately represent and predict costs in situations where incremental changes occur. By the end of this article, readers will be equipped with the knowledge and tools to solve similar problems involving arithmetic progressions and cost calculations. They will also appreciate the power of mathematics in modeling real-world scenarios and making informed decisions based on quantitative analysis.

Understanding the Problem

The problem states that the cost of digging a well increases after every meter. The initial cost for digging the first meter is ₹150, and this cost rises by ₹50 for each subsequent meter. This pattern forms an arithmetic progression, where the first term (a) is 150 and the common difference (d) is 50. The essence of understanding this problem lies in recognizing this arithmetic progression and applying its properties to calculate the cost at different depths. An arithmetic progression is a sequence of numbers in which the difference between any two consecutive terms is constant. This constant difference is known as the common difference. In our case, the sequence of costs forms an arithmetic progression: ₹150 for the first meter, ₹200 for the second meter, ₹250 for the third meter, and so on. The consistent increase of ₹50 makes this progression arithmetic. To effectively solve the problem, we need to identify the key elements: the first term (₹150), the common difference (₹50), and the number of terms (meters dug). Once these elements are known, we can use the formulas for arithmetic progressions to find the cost for digging any particular meter or the total cost for digging a certain depth. Furthermore, understanding the problem involves recognizing that the cost for each additional meter is independent of the cost of the previous meters. This means that the cost for digging the tenth meter, for example, is only dependent on the initial cost and the common difference, not on the costs of the first nine meters. This independence allows us to use the arithmetic progression formulas directly without needing to sum up the costs for each individual meter. In summary, understanding the problem requires identifying the arithmetic progression, recognizing the first term and common difference, and understanding how these elements can be used to calculate costs for individual meters or total depths.

Deriving the Formula for the Cost of Digging

To determine the cost of digging a well at any given meter, we need to derive a formula based on the arithmetic progression. The formula for the nth term (an) of an arithmetic progression is given by: an = a + (n - 1)d, where a is the first term, n is the term number (in this case, the meter number), and d is the common difference. In our scenario, a = ₹150 (the cost for the first meter) and d = ₹50 (the increase in cost for each subsequent meter). Substituting these values into the formula, we get: an = 150 + (n - 1)50. This formula allows us to calculate the cost for digging any specific meter. For instance, if we want to find the cost for digging the 5th meter, we would substitute n = 5 into the formula: a5 = 150 + (5 - 1)50 = 150 + 4 * 50 = 150 + 200 = ₹350. Therefore, the cost for digging the 5th meter is ₹350. Similarly, if we want to calculate the total cost for digging up to a certain depth, we can use the formula for the sum of the first n terms of an arithmetic progression: Sn = n/2 [2a + (n - 1)d]. This formula gives us the total cost for digging n meters. For example, to find the total cost for digging 10 meters, we would substitute n = 10, a = 150, and d = 50 into the formula: S10 = 10/2 [2 * 150 + (10 - 1)50] = 5 [300 + 9 * 50] = 5 [300 + 450] = 5 * 750 = ₹3750. Therefore, the total cost for digging 10 meters is ₹3750. By deriving and understanding these formulas, we can efficiently calculate the cost for digging at any depth and make informed decisions regarding the financial aspects of well-digging projects. These formulas also provide a foundation for more complex cost analyses, such as determining the most cost-effective depth to dig based on water availability and financial constraints.

Calculating the Cost for Different Depths

Using the derived formulas, we can now calculate the cost for digging at various depths. This involves applying both the formula for the cost of a specific meter (an = 150 + (n - 1)50) and the formula for the total cost of digging up to a certain depth (Sn = n/2 [2a + (n - 1)d]). Let's consider some examples to illustrate this. First, let's calculate the cost for digging the 8th meter. Using the formula an = 150 + (n - 1)50, we substitute n = 8: a8 = 150 + (8 - 1)50 = 150 + 7 * 50 = 150 + 350 = ₹500. So, the cost for digging the 8th meter is ₹500. Next, let's calculate the total cost for digging 15 meters. Using the formula Sn = n/2 [2a + (n - 1)d], we substitute n = 15, a = 150, and d = 50: S15 = 15/2 [2 * 150 + (15 - 1)50] = 7.5 [300 + 14 * 50] = 7.5 [300 + 700] = 7.5 * 1000 = ₹7500. Therefore, the total cost for digging 15 meters is ₹7500. We can also calculate the cost for a range of depths to create a cost table, which can be useful for budgeting and planning. For example, we can calculate the costs for digging 5, 10, 15, and 20 meters and present them in a table. This allows for a quick comparison of costs at different depths. Furthermore, these calculations can be extended to more complex scenarios, such as determining the depth at which the cost exceeds a certain budget or comparing the cost of digging with other water sourcing options. By applying these formulas and performing various calculations, we gain a comprehensive understanding of the cost implications at different depths and can make informed decisions based on our specific needs and constraints. This practical application of arithmetic progressions demonstrates their usefulness in real-world cost analysis and financial planning.

Practical Applications and Implications

The concept of increasing costs with depth, as demonstrated in the well-digging problem, has numerous practical applications and implications beyond just this specific scenario. Understanding how costs escalate with each increment is crucial in various fields, including construction, mining, and even subscription-based services. In construction, for example, the cost of laying foundations often increases with depth due to the need for more materials, labor, and specialized equipment. Similarly, in mining, the cost of extracting resources rises as the mine gets deeper, necessitating more sophisticated machinery and safety measures. This principle is also evident in subscription-based services, where the cost may increase with the number of features or users added. For instance, a software subscription might start at a base price but increase as more users are included or additional functionalities are activated. Understanding these cost implications is essential for effective budgeting and financial planning. By recognizing the arithmetic progression in cost increases, businesses and individuals can accurately forecast expenses and make informed decisions about resource allocation. This knowledge can help in determining the optimal depth for a well, the feasibility of a mining project, or the most cost-effective subscription plan for a software service. Furthermore, this concept highlights the importance of considering long-term costs versus short-term gains. While digging deeper or adding more features may seem beneficial initially, the escalating costs can quickly outweigh the advantages if not properly managed. Therefore, a thorough cost-benefit analysis is crucial in any scenario where costs increase incrementally. In addition to financial planning, understanding the implications of increasing costs can also influence project management and resource optimization. By identifying the factors that contribute to cost escalation, project managers can implement strategies to mitigate these increases and ensure the project stays within budget. This might involve using more efficient technologies, optimizing labor allocation, or renegotiating contracts with suppliers. In conclusion, the principle of increasing costs with depth, as illustrated by the well-digging problem, has far-reaching applications across various industries and sectors. By understanding and applying this concept, individuals and organizations can make more informed financial decisions, optimize resource allocation, and effectively manage costs in a wide range of scenarios.

Conclusion

In conclusion, the problem of calculating the cost of digging a well, where the price increases after every meter, provides an excellent illustration of arithmetic progressions and their practical applications. We have seen how the cost structure forms an arithmetic progression with a fixed initial cost and a consistent increase for each subsequent meter. By understanding the properties of arithmetic progressions, we were able to derive formulas for calculating the cost at any given depth and the total cost for digging up to a certain depth. These formulas, an = 150 + (n - 1)50 for the cost of the nth meter and Sn = n/2 [2a + (n - 1)d] for the total cost of digging n meters, allow for efficient and accurate cost calculations. Through various examples, we demonstrated how these formulas can be applied to determine the cost for individual meters and the cumulative cost for different depths. This practical application highlights the relevance of mathematical concepts in real-world scenarios and their usefulness in financial planning and decision-making. Moreover, we discussed the broader implications of this concept, noting that increasing costs with depth or quantity is a common phenomenon in various industries, including construction, mining, and subscription services. Understanding these cost escalations is crucial for effective budgeting, resource allocation, and project management. The ability to accurately forecast costs and make informed decisions based on quantitative analysis is a valuable skill in both personal and professional contexts. In summary, this exploration of the well-digging cost problem not only reinforces our understanding of arithmetic progressions but also underscores the importance of mathematical literacy in navigating and solving everyday challenges. By applying mathematical principles to practical situations, we can gain valuable insights and make more informed decisions, ultimately leading to better outcomes.