Assessment Instructions: Tech Company Development Scenario

Assessment Instructions Scenario A tech company has developed a new comp

Assessment Instructions Scenario A tech company has developed a new compact, high efficiency battery for hand-held devices. Market projections have estimated the cost and revenue of manufacturing these batteries by the equations graphed below. Show and explain all steps in your responses to the following parts of the assignment using the Algebra concepts discussed within the course. All mathematical steps and explanations must be typed up and formatted using the equation editor. Part 1 : Use the substitution method to determine the point where the cost equals the revenue. Part 2 : Interpret your results from Part 1 in the context of the problem. Part 3 : Do your results from Part 1 correspond with the graph? Explain. Part 4 : Profit is found by subtracting cost from revenue. Write an equation in the same variables to represent the profit. Part 5 : Find the profit from producing 100 thousand batteries.

Paper For Above instruction

The scenario presents a technological innovation by a company that has developed a high-efficiency, compact battery for handheld devices. The key focus of this analysis involves algebraic techniques to determine the breakeven point (where cost equals revenue), interpret what this point signifies for the company, relate the mathematical findings to the graphical data, establish a profit equation, and calculate the profit for a specific production quantity. This comprehensive analysis not only demonstrates the practical application of algebra but also offers insights into the company's financial planning and decision-making process.

Introduction

In the competitive field of portable electronics, efficiency and cost-effectiveness are crucial. The company's recent development of a new battery aims to meet market demands while optimizing production costs and revenue. Understanding the relationship between these financial factors through algebraic equations enables decision-makers to identify optimal production levels and assess financial viability. This paper employs algebraic methods—including substitution—to analyze the given cost and revenue functions, interpret the results, and apply them to real-world manufacturing scenarios.

Part 1: Using the substitution method to find the breakeven point

Assume the cost function is defined by C(x), and the revenue function by R(x), where x represents the number of batteries produced (in thousands). The problem provides specific equations for cost and revenue, which are presumably given in the form of graphs. For the sake of analysis, consider typical equations such as:

  • Cost function: C(x) = a constant + variable cost per unit × x
  • Revenue function: R(x) = price per unit × x

Suppose, for example, C(x) = 200 + 3x and R(x) = 5x, which exemplify the cost and revenue functions based on the projected data. To find the breakeven point using the substitution method, set C(x) equal to R(x):

200 + 3x = 5x

Subtract 3x from both sides:

200 = 2x

Solve for x:

x = 100

Thus, the breakeven point occurs when 100 thousand batteries are produced.

Part 2: Interpretation of the results

This result indicates that at a production volume of 100,000 batteries, the company's total revenue will exactly cover its total costs, resulting in zero profit or loss. Producing fewer than 100,000 units would not cover costs, leading to a loss, whereas producing more than 100,000 units would yield profit. Such information is vital for strategic planning, resource allocation, and setting production targets to ensure financial sustainability.

Part 3: Correspondence with the graph and explanation

The algebraic result should align with the graphical analysis, where the cost and revenue curves intersect at x = 100. On the graph, the point of intersection visually confirms the algebraic solution. If the graph shows the cost and revenue lines crossing at x = 100, it affirms the correctness of the algebraic process. Discrepancies might arise due to graphical scale or inaccuracies in the graph, but generally, the algebraic solution and graphical intersection should coincide, validating both methods.

Part 4: Formulating the profit equation

Profit (P) is the difference between revenue and cost:

P(x) = R(x) - C(x)

Using the example functions:

P(x) = 5x - (200 + 3x) = 5x - 200 - 3x = 2x - 200

This equation models profit as a function of production volume. When x = 100, profit becomes:

P(100) = 2(100) - 200 = 200 - 200 = 0

Part 5: Calculating profit for 100 thousand batteries

Substituting x = 100 into the profit equation:

P(100) = 2(100) - 200 = 200 - 200 = 0

Therefore, producing 100,000 batteries results in neither profit nor loss, confirming the breakeven point calculated earlier.

Conclusion

This analysis demonstrates how algebraic methods, including substitution, provide critical insights into the financial aspects of new product development. By pinpointing the breakeven production volume, interpreting its significance, and relating it to graphical data, companies can make informed strategic decisions. The profit equation further facilitates scenario analysis, ensuring that manufacturing plans align with profitability goals. Overall, the algebraic approach serves as a valuable tool in financial and operational planning within the technology manufacturing sector.

References

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