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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?

Short Answer

Expert verified
Max profit is 445k at 12.5k widgets; breakeven at 1.95k. Zero widgets loss 180k; 15k widgets yield 420k profit.

Step by step solution

01

Graphing the Profit Function

The given function for the yearly profit is \( P = -180 + 100n - 4n^2 \). To graph this function, choose a range of values for \( n \) (from 0 to 20) and compute the corresponding \( P \) values. Plot these points on a graph with \( n \) on the x-axis and \( P \) on the y-axis. This will form a downward-facing parabola since the coefficient of the \( n^2 \) term is negative.
02

Calculate P(0)

Substitute \( n = 0 \) into the function to calculate \( P(0) \):\[ P(0) = -180 + 100(0) - 4(0)^2 = -180. \]This means that if no widgets are sold, the company incurs a loss of \(-180\) thousand dollars, due to fixed costs.
03

Calculate Profit for 15 Thousand Widgets

Substitute \( n = 15 \) into the function to find the profit when 15 thousand widgets are sold:\[ P(15) = -180 + 100(15) - 4(15)^2 = -180 + 1500 - 900 = 420. \]Hence, the profit is 420 thousand dollars.
04

Approximate the Break-even Point

To find the break-even points, set \( P = 0 \) and solve for \( n \):\[ 0 = -180 + 100n - 4n^2. \]Rearrange to form a quadratic equation:\[ 4n^2 - 100n + 180 = 0. \]Use the quadratic formula \[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] with \( a = 4, b = -100, c = 180 \):\[ n = \frac{-(-100) \pm \sqrt{(-100)^2 - 4(4)(180)}}{2(4)} = \frac{100 \pm \sqrt{10000 - 2880}}{8}. \]Simplify:\[ n = \frac{100 \pm \sqrt{7120}}{8} \approx \frac{100 \pm 84.4}{8}. \]Calculate the roots:\[ n_1 \approx \frac{184.4}{8} \approx 23.05 \text{ (not valid, beyond range)}, \]\[ n_2 \approx \frac{15.6}{8} \approx 1.95. \]The valid break-even point within the range is approximately \( n = 1.95 \) thousand widgets.
05

Determine the Maximum Profit

The maximum profit occurs at the vertex of the parabola, which can be found using \( n = -\frac{b}{2a} \). Here, \( a = -4 \) and \( b = 100 \):\[ n = -\frac{100}{2(-4)} = \frac{100}{8} = 12.5. \]Substitute this back into the profit equation:\[ P(12.5) = -180 + 100(12.5) - 4(12.5)^2 = 445. \]The maximum profit is \( 445 \) thousand dollars.

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Key Concepts

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

Quadratic Equations
Quadratic equations are a type of polynomial equation featuring a variable raised to the power of two. These equations are typically written in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. In the context of the given profit function, \( P = -180 + 100n - 4n^2 \), the term \(-4n^2 \) reveals that this is a quadratic equation. Here, \( a = -4 \), \( b = 100 \), and \( c = -180 \). The quadratic nature results in the graph of the profit function forming a parabola.

Why is this significant? Quadratic equations are crucial in various real-world applications, from physics to finance. In this case, it helps model the profit based on the number of widgets sold. Solutions to these equations often provide key insights, such as determining break-even points and maximum or minimum values.

One hallmark of quadratic equations is their two solutions or roots, which can be found using various methods such as factoring, completing the square, or the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). The solution for \( n \) reveals the break-even points where profit equals zero.
Graphing Techniques
Graphing techniques for quadratic functions are essential for visualizing relationships within equations. For a quadratic function, the graph is a parabola - a symmetrical curve. Consider the profit function \( P = -180 + 100n - 4n^2 \), where the graph slopes downward, forming a parabola due to the negative coefficient of the \( n^2 \) term.

To visualize this curve effectively:
  • Choose a range of values for \( n \), such as from 0 to 20 in this case, reflecting the valid domain for widgets sold in thousands.
  • Calculate the corresponding \( P \) values.
  • Plot these \( n, P \) pairs on a coordinate plane with \( n \) on the x-axis and \( P \) on the y-axis.
By graphing these points, a clear "U" shaped curve will form. This enables visualization of key points such as where the curve crosses the x-axis (indicating break-even points) and the highest point of the parabola (indicating maximum profit).

Graphing thus provides a powerful tool for understanding the range within which the company shifts from loss to profit and when it yields its peak financial performance.
Break-even Analysis
Break-even analysis is a fundamental concept in business, used to determine when a company will start making a profit. In the realm of quadratic functions, this analysis involves finding where the equation equals zero. For the profit function \( P = -180 + 100n - 4n^2 \), finding \( n \) when \( P = 0 \) gives the break-even points.

To conduct this analysis, set the equation \[ 0 = -180 + 100n - 4n^2 \] and simplify using the quadratic formula. This process yields:
  • \( n_1 \approx 23.05 \), which lies outside the viable range.
  • \( n_2 \approx 1.95 \), which means selling approximately 1,950 widgets is the break-even point within the valid range.
At this point, the producer covers all costs, both fixed and variable, thus incurring neither profit nor loss.

Understanding this threshold informs business decisions, such as pricing strategy and production levels, to ensure profitability.
Vertex of a Parabola
The vertex of a parabola represents either the maximum or minimum value the function can achieve. In the context of a downward-facing parabola, like our profit function \( P = -180 + 100n - 4n^2 \), the vertex corresponds to the point of maximum profit.

To find the vertex, utilize the formula \( n = -\frac{b}{2a} \). Here:
  • \( b = 100 \)
  • \( a = -4 \)
Solving gives \( n = 12.5 \), meaning at 12,500 widgets sold, the profit is maximized.

Plugging \( n = 12.5 \) back into the original equation calculates the maximum profit:\[ P(12.5) = -180 + 100(12.5) - 4(12.5)^2 = 445. \]This result informs stakeholders of the peak financial performance achievable and helps in planning production and sales strategies to align with this optimal level.

Thus, the vertex not only marks the highest point on the graph but also serves as a critical target for operational planning and resource allocation.

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Most popular questions from this chapter

An important model for commercial fisheries is that of Beverton and Holt. \({ }^{31}\) It begins with the study of a single cohort of fishthat is, all the fish in the study are born at the same time. For a cohort of the North Sea plaice (a type of flatfish), the number \(N=N(t)\) of fish in the population is given by $$ N=1000 e^{-0.1 t} $$ and the weight \(w=w(t)\) of each fish is given by $$ w=6.32\left(1-0.93 e^{-0.095 t}\right)^{3} $$ Here \(w\) is measured in pounds and \(t\) in years. The variable \(t\) measures the so-called recruitment age, which we refer to simply as the age. The biomass \(B=B(t)\) of the fish cohort is defined to be the total weight of the cohort, so it is obtained by multiplying the population size by the weight of a fish. a. If a plaice weighs 3 pounds, how old is it? b. Use the formulas for \(N\) and \(w\) given above to find a formula for \(B=B(t)\), and then make a graph of \(B\) against \(t\). (Include ages through 20 years.) c. At what age is the biomass the largest? d. In practice, fish below a certain size can't be caught, so the biomass function becomes relevant only at a certain age. i. Suppose we want to harvest the plaice population at the largest biomass possible, but a plaice has to weigh 3 pounds before we can catch it. At what age should we harvest? ii. Work part i under the assumption that we can catch plaice weighing at least 2 pounds.

The profit \(P\), in thousands of dollars, that a manufacturer makes is a function of the number \(N\) of items produced in a year, and the formula is $$ P=-0.2 N^{2}+3.6 N-9 . $$ a. Express using functional notation the profit at a production level of 5 items per year, and then calculate that value. b. Determine the two break-even points for this manufacturer-that is, the two production levels at which the profit is zero. c. Determine the maximum profit if the manufacturer can produce at most 20 items in a year.

A breeding group of foxes is introduced into a protected area, and the population growth follows a logistic pattern. After \(t\) years the population of foxes is given by $$ N=\frac{37.5}{0.25+0.76^{t}} \text { foxes } $$ a. How many foxes were introduced into the protected area? b. Make a graph of \(N\) versus \(t\) and explain in words how the population of foxes increases with time. c. When will the fox population reach 100 individuals?

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 \(\$ 1200\) per month, and the variable cost is \(\$ 40\) per widget (so it costs \(\$ 40\) to produce 1 widget). Let \(N\) be the number 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, of a widget at which \(N\) widgets can be sold is given by the formula \(p=53-0.01 N\). 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 here that the manufacturer produces the widgets in blocks of 50 , so a table setup showing \(N\) in multiples of 50 is appropriate. e. Use your formula from part c to determine the production level at which profit is maximized if the manufacturer can produce at most 1500 widgets in a month. As in part d, assume that the manufacturer produces the widgets in blocks of 50 .

In a study of a marine tubeworm, scientists developed a model for the time \(T\) (measured in years) required for the tubeworm to reach a length of \(L\) meters. \({ }^{23}\) On the basis of their model, they estimate that 170 to 250 years are required for the organism to reach a length of 2 meters, making this tubeworm the longest-lived noncolonial marine invertebrate known. The model is $$ T=\frac{20 e^{b L}-28}{b} $$ Here \(b\) is a constant that requires estimation. What is the value of \(b\) if the lower estimate of 170 years to reach a length of 2 meters is correct? What is the value of \(b\) if the upper estimate of 250 years to reach a length of 2 meters is correct?

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