/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 The background for this exercise... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 11

The background for this exercise can be found in Exercises 11, 12,13, and 14 in Section 1.4. A manufacturer of widgets has fixed costs of \(\$ 600\) per month, and the variable cost is \(\$ 60\) per thousand widgets (so it costs \(\$ 60\) to produce a thousand widgets). Let \(N\) be the number, in thousands, of widgets produced in a month. a. Find a formula for the manufacturer's total cost \(C\) as a function of \(N\). b. The highest price \(p\), in dollars per thousand widgets, at which \(N\) can be sold is given by the formula \(p=70-0.03 N\). Using this, find a formula for the total revenue \(R\) as a function of \(N\). c. Use your answers to parts a and \(b\) to find a formula for the profit \(P\) of this manufacturer as a function of \(N\). d. Use your formula from part c to determine the production level at which profit is maximized if the manufacturer can produce at most 300 thousand widgets in a month.

Short Answer

Expert verified
The manufacturer should produce approximately 167 thousand widgets for maximum profit.

Step by step solution

01

Fixed and Variable Costs

The total cost \(C\) for producing \(N\) thousands of widgets is the sum of fixed costs and variable costs. The fixed cost is \$600 per month. The variable cost for \(N\) thousands of widgets is \(60N\). Therefore, the total cost \(C\) can be formulated as: \[ C = 600 + 60N \]
02

Calculate Revenue

The revenue \(R\) is calculated by multiplying the price \(p\) at which the widgets are sold by the number of thousands of widgets \(N\) produced. From the formula, \(p = 70 - 0.03N\) dollars per thousand widgets, the total revenue can be expressed as: \[ R = N(70 - 0.03N) = 70N - 0.03N^2 \]
03

Calculate Profit

Profit \(P\) is determined by subtracting the total cost \(C\) from the total revenue \(R\). Using the expressions for \(C\) and \(R\) derived earlier, profit can be formulated as:\[ P = R - C = (70N - 0.03N^2) - (600 + 60N) = 10N - 0.03N^2 - 600 \]
04

Determine Maximum Profit

To maximize profit \(P\), find the value of \(N\) that maximizes the quadratic function \(P = -0.03N^2 + 10N - 600\). The maximum of a quadratic function \(ax^2 + bx + c\) occurs at \(x = -\frac{b}{2a}\). In this case, \(a = -0.03\) and \(b = 10\), so:\[ N = -\frac{10}{2(-0.03)} = \frac{10}{0.06} = 166.67 \]Since the manufacturer can produce at most 300 thousand widgets, the production level that maximizes profit is approximately 167 thousand widgets.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Function Optimization
Optimization, especially of quadratic functions, is a pivotal concept in economics and business for finding maximum or minimum values. In this exercise, we want to find the number of widgets that should be produced to maximize profit. The profit function we need to optimize is a quadratic function given by:\[ P = -0.03N^2 + 10N - 600 \]The vertex formula for a quadratic equation \( ax^2 + bx + c \) helps in finding the optimal value of \( x \), where \( x = -\frac{b}{2a} \). Here, \( a = -0.03 \) and \( b = 10 \). To find the maximum profit, plug these values into the formula:\[ N = -\frac{10}{2 \times -0.03} = 166.67 \]Since widgets are produced in whole numbers and the manufacturer can only produce up to 300, rounding to 167 maximizes profit. Quadratic optimization is a robust tool for economic decision-making, allowing businesses to efficiently balance production levels with market demand.
Cost Function
The cost function is essential for understanding overall expenses in widget production. It comprises both fixed and variable costs, providing a clear picture of manufacturing outlays.- **Fixed Costs:** These are expenses that do not change with production levels, such as rent or salaries. For the widget manufacturer, the fixed cost is \( \\(600 \).- **Variable Costs:** These are proportional to the production level. In this case, the cost is \( \\)60 \) per thousand widgets, represented as \( 60N \), where \( N \) denotes the number of thousands of widgets.Combining these, the total cost \( C \) is:\[ C = 600 + 60N \]Understanding this cost function helps manufacturers budget effectively, ensuring not only coverage of expenses but also potential profit realization.
Revenue Function
The revenue function is crucial, defining the income from selling widgets. Calculating revenue involves understanding how selling price and quantity interact.- **Price per Thousand Widgets:** The price \( p \) decreases with more widgets supplied, given as \( p = 70 - 0.03N \).- **Revenue Calculation:** Revenue \( R \) is the product of this price and the number of widgets sold: \[ R = N(70 - 0.03N) = 70N - 0.03N^2 \]This formula reflects typical market behavior where higher production reduces prices but can boost total revenue within certain ranges. Businesses can thus strategically adjust output to maximize revenue while considering market demand.
Widget Manufacturing Economics
Widget manufacturing involves not just physical production but also complex economic decisions. This includes understanding cost, revenue, and profit interrelations to make informed choices:In widget manufacturing, the intersection of these financial factors shapes profitability:- **Profit Consideration:** With profit defined as \( P = R - C \), it calculates as: \[ P = (70N - 0.03N^2) - (600 + 60N) \] Leading to: \[ P = 10N - 0.03N^2 - 600 \]Effective economics management involves balancing production costs against revenue capabilities. For instance, determining optimal widget quantities ensures the highest possible profit under production constraints. Understanding these components is vital for achieving cost-effectiveness and financial sustainability in manufacturing.

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