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Chapter 2: Problem 17

The background for this exercise can be found in Exercises 11, 12, 13, and \(14 \mathrm{in} \mathrm{Sec}\) tion 1.4. A manufacturer of widgets has fixed costs of \(\$ 700\) per month, and the variable cost is \(\$ 65\) per thousand widgets (so it costs \(\$ 65\) to produce 1 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 \(p=75-0.02 \mathrm{~N}\). 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 two break-even points for this manufacturer. Assume that the manufacturer can produce at most 500 thousand widgets in a month.

Short Answer

Expert verified
The formulas are: Total cost \( C = 700 + 65N \); Revenue \( R = 75N - 0.02N^2 \); Profit \( P = 10N - 0.02N^2 - 700 \). Break-even points are at \( N \approx 100 \) and \( N \approx 350 \).

Step by step solution

01

Formula for Total Cost

The total cost \( C \) consists of fixed costs and variable costs. The fixed cost is \\(700 per month, and the variable cost is \\)65 per thousand widgets. If \( N \) is the number of thousand widgets produced, the formula for the total cost is given by:\[ C = 700 + 65N \] This formula incorporates both the fixed and variable components of the cost.
02

Formula for Total Revenue

The price per thousand widgets is given by the formula \( p = 75 - 0.02N \). The total revenue \( R \) from selling \( N \) thousand widgets is calculated by multiplying the price \( p \) by the number of thousand widgets \( N \). Therefore, \[ R = p \times N = (75 - 0.02N)N \] Expanding this, we get:\[ R = 75N - 0.02N^2 \] This formula represents the total revenue as a function of \( N \).
03

Formula for Profit

The profit \( P \) is the difference between revenue \( R \) and cost \( C \). Using the formulas derived in Steps 1 and 2, we have:\[ P = R - C = (75N - 0.02N^2) - (700 + 65N) \] Simplifying this, we obtain:\[ P = 75N - 0.02N^2 - 700 - 65N \]\[ P = 10N - 0.02N^2 - 700 \]This formula represents the profit as a function of \( N \).
04

Finding Break-even Points

Break-even points occur when the profit \( P \) equals zero. Set \( P = 0 \) using the formula from Step 3:\[ 0 = 10N - 0.02N^2 - 700 \]Rearrange it into a standard quadratic equation:\[ 0.02N^2 - 10N + 700 = 0 \]Using the quadratic formula \( N = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 0.02 \), \( b = -10 \), and \( c = 700 \), solve for \( N \):\[ N = \frac{10 \pm \sqrt{(-10)^2 - 4 \times 0.02 \times 700}}{0.04} \]\[ N = \frac{10 \pm \sqrt{100 - 56}}{0.04} \]\[ N = \frac{10 \pm \sqrt{44}}{0.04} \]Calculating the square root and solving gives two break-even points \( N \approx 100 \) and \( N \approx 350 \).Check these points to ensure that they lie within the maximum production constraint of 500 thousand widgets.

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Key Concepts

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

Profit Maximization
Profit maximization involves determining the level of production that allows a company to achieve the highest possible profit. In this exercise, the manufacturer aims to determine the optimal number of thousands of widgets ( \(N\)) to produce in order to maximize profits. Profit ( \(P\)) is calculated as the total revenue ( \(R\)) minus total costs ( \(C\)). For our widgets manufacturer, the profit formula is derived from:
  • Revenue formula: \(R = 75N - 0.02N^2\), which accounts for how revenue depends on both the number of widgets sold and the decreasing price per widget as output increases.
  • Cost formula: \(C = 700 + 65N\), representing both fixed and variable costs.
Plugging these into the profit equation gives:\[P = 10N - 0.02N^2 - 700\]To maximize profit, one would typically take the derivative of this profit function with respect to \(N\), set it equal to zero, and solve for \(N\). However, the exercise focuses on finding where profit becomes zero (break-even), not maximization.
Break-even Analysis
Break-even analysis is essential for businesses to understand when they will start making a profit. It calculates the point at which revenue equals total costs, resulting in neither profit nor loss. For the widget manufacturer, the break-even analysis involves finding values of \(N\) where profit \(P\) is zero. Using the profit formula:\[P = 10N - 0.02N^2 - 700\]Set this equation to zero and solve for \(N\):\[0 = 10N - 0.02N^2 - 700\]Rearrange into a quadratic equation:\[0.02N^2 - 10N + 700 = 0\]Solving this will provide the break-even points, or the number of thousand widgets needed to break even. The solutions are found using the quadratic formula.
Quadratic Formula
The quadratic formula is a powerful mathematical tool used to find the roots of a quadratic equation. In the context of this problem, the formula helps locate the production levels where the widget manufacturer neither gains nor loses money. The quadratic equation presented is:\[0.02N^2 - 10N + 700 = 0\]This equation matches the standard form \(ax^2 + bx + c = 0\), where \(a\) is 0.02, \(b\) is -10, and \(c\) is 700. The quadratic formula is:\[N = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Plugging in the values, we find:\[N = \frac{10 \pm \sqrt{(-10)^2 - 4 \times 0.02 \times 700}}{0.04}\]Solving the square root and the equation provides two break-even points: \(N \approx 100\) and \(N \approx 350\). These points indicate the production levels where profits are zero, within the maximum capacity of 500 thousand widgets allowed per month.

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Most popular questions from this chapter

The yearly profit \(P\) for a widget producer is a function of the number \(n\) of widgets sold. The formula is $$ P=-180+100 n-4 n^{2} . $$ Here \(P\) is measured in thousands of dollars, \(n\) is measured in thousands of widgets, and the formula is valid up to a level of 20 thousand widgets sold. a. Make a graph of \(P\) versus \(n\). b. Calculate \(P(0)\) and explain in practical terms what your answer means. c. What profit will the producer make if 15 thousand widgets are sold? d. The break-even point is the sales level at which the profit is 0 . Approximate the break-even point for this widget producer. e. What is the largest profit possible?

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