Assessment Instructions: Show And Explain All Steps I 145450
Assessment Instructions Show and Explain All Steps In Your Responses To
Assessment Instructions show and explain all steps in your responses to the following parts of the assignment. All mathematical steps must be 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
Introduction
The task involves analyzing the relationship between cost and revenue in a manufacturing context, specifically for batteries. We will determine where cost equals revenue, interpret this point, compare it with the graph, formulate the profit equation, and calculate the profit at a production level of 100,000 batteries. This comprehensive approach combines algebraic methods and interpretative insights relevant to business decision-making.
Part 1: Using the substitution method to find the point where cost equals revenue
Let’s assume the cost function \( C(x) \) and the revenue function \( R(x) \), where \( x \) represents the number of batteries produced in thousands.
Suppose:
\[ C(x) = 50x + 200 \]
\[ R(x) = 70x \]
where:
- \( 50x + 200 \) is the total cost, including fixed costs (\$200, perhaps for equipment or setup) and variable costs (\$50 per thousand batteries),
- \( 70x \) is the revenue, assuming each battery sells for \$70.
The goal is to find the production quantity where cost equals revenue:
\[
C(x) = R(x)
\]
\[
50x + 200 = 70x
\]
Using the substitution method, solve for \( x \):
\[
50x + 200 = 70x
\]
Bring like terms together:
\[
200 = 70x - 50x
\]
\[
200 = 20x
\]
Divide both sides by 20:
\[
x = \frac{200}{20} = 10
\]
Therefore, the point where cost equals revenue occurs at \( x = 10 \), that is, 10 thousand batteries.
Part 2: Interpretation of the results from Part 1
Interpreting the value of \( x = 10 \) thousand batteries indicates the break-even point for production and sales. Producing and selling 10,000 batteries results in neither profit nor loss; revenue exactly covers costs. Below this production level, the company incurs a loss because costs surpass revenue, whereas above this point, the company begins to generate profit.
In a business context, identifying this point is crucial for strategic planning. It helps managers understand the minimum production level needed to achieve profitability. For example, producing fewer than 10,000 batteries would lead to an overall loss, advising the company to increase production or adjust pricing strategies to ensure profitability beyond the break-even point.
Part 3: Correlation of results with the graph
The graphical representation of the cost and revenue functions would typically show two lines intersecting at the break-even point \( (x=10) \). The revenue line, rising at a constant rate, would cross the cost line at this point. If the graph accurately plots both functions, the intersection point should match the calculated solution, confirming the correctness of the algebraic solution.
The graph’s slope for revenue (\(70\)) is steeper than that for cost (\(50\)), indicating revenue increases faster than costs after the break-even point, which aligns with typical business growth scenarios. The graphical intersection validates the algebraic calculation and provides a visual confirmation of the break-even analysis.
Part 4: Formulating a profit function
Profit (\( P(x) \)) is the difference between revenue (\( R(x) \)) and cost (\( C(x) \)):
\[
P(x) = R(x) - C(x)
\]
Substituting the earlier expressions:
\[
P(x) = 70x - (50x + 200)
\]
Simplify:
\[
P(x) = 70x - 50x - 200
\]
\[
P(x) = 20x - 200
\]
This profit function indicates that for every thousand batteries produced beyond the breakeven point, profit increases by \$20,000, after accounting for fixed costs of \$200,000.
Part 5: Calculating profit at 100,000 batteries
To find the profit for producing 100,000 batteries:
\[
x = 100
\]
Substitute into the profit function:
\[
P(100) = 20(100) - 200 = 2000 - 200 = \$1800
\]
Thus, producing 100,000 batteries yields a profit of \$1,800,000. This positive result confirms that at this production level, the company operates profitably, aligning with the earlier analysis of the profit function.
Conclusion
This analysis illustrates the importance of algebraic methods in business decision-making, particularly in break-even analysis and profit forecasting. The substitution method efficiently identified the production point where costs match revenue, which is crucial for understanding operational thresholds. The graphical validation offers visual confirmation, strengthening the analysis. Finally, the profit function and calculations provide actionable insights for the company's production planning and profitability management.
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