/*! 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 7 The cost of recycling \(q\) tons... [FREE SOLUTION] | 91Ó°ÊÓ

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The cost of recycling \(q\) tons of paper is given in the fol. lowing table. Estimate the marginal cost at \(q=2000\) Give units and interpret your answer in terms of cost. At approximately what production level does marginal cost appear smallest?$$\begin{array}{c|c|c|c|c|c|c} \hline q \text { (tons) } & 1000 & 1500 & 2000 & 2500 & 3000 & 3500 \\ \hline C(q) \text { (dollars) } & 2500 & 3200 & 3640 & 3825 & 3900 & 4400 \\ \hline \end{array}$$

Short Answer

Expert verified
The marginal cost at \( q=2000 \) is approximately \( 0.625 \) dollars per ton. The smallest marginal cost occurs at \([2500, 3000]\) with \( 0.15 \) dollars per ton.

Step by step solution

01

Define Marginal Cost

The marginal cost is the rate of change of cost with respect to quantity, essentially the derivative \( C'(q) \). It can be estimated using the change in cost over the change in quantity \( \Delta q \).
02

Identify Intervals

To estimate the marginal cost at \( q=2000 \), we can look at the interval \( [1500, 2000] \) and \( [2000, 2500] \) in the table, where \( q = 2000 \) is situated.
03

Calculate Marginal Cost in First Interval

For the interval \( [1500, 2000] \), use the formula: \( \frac{C(2000) - C(1500)}{2000 - 1500} = \frac{3640 - 3200}{500} = \frac{440}{500} = 0.88 \) dollars per ton.
04

Calculate Marginal Cost in Second Interval

For the interval \( [2000, 2500] \), use the formula: \( \frac{C(2500) - C(2000)}{2500 - 2000} = \frac{3825 - 3640}{500} = \frac{185}{500} = 0.37 \) dollars per ton.
05

Average Marginal Cost Estimate at q=2000

Estimate the marginal cost at \( q=2000 \) by averaging the values from the intervals: \( \frac{0.88 + 0.37}{2} = 0.625 \) dollars per ton. Thus, the marginal cost at \( q=2000 \) is approximately \( 0.625 \) dollars per ton.
06

Determine Smallest Marginal Cost

Calculate the marginal costs between all consecutive intervals: \([1000, 1500]\), \([1500, 2000]\), \([2000, 2500]\), \([2500, 3000]\), \([3000, 3500]\), and find the smallest.\([1000, 1500]\): \( \frac{3200 - 2500}{500} = 1.40 \) dollars per ton.\([2500, 3000]\): \( \frac{3900 - 3825}{500} = 0.15 \) dollars per ton.\([3000, 3500]\): \( \frac{4400 - 3900}{500} = 1.00 \) dollars per ton.\ The smallest marginal cost is \( 0.15 \) dollars per ton at interval \([2500, 3000]\).

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Key Concepts

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

calculus
Calculus acts as the foundational mathematical tool for studying changes, making it vital in scenarios like economic analysis, growth trends, and engineering designs. Among its many applications, it plays an essential role in understanding change rates, a concept also central to estimating marginal costs. In business and economics, knowing how a cost changes as production levels shift is crucial. This change is called perhaps the marginal cost.

Calculus allows us to delve into these change rates by using derivatives, which are calculated by finding the limit of the average rate of change over increasingly smaller intervals. This mathematical discipline helps businesses and policymakers determine how variables such as cost and production interact, ultimately enabling them to make more informed decisions about resource allocation and pricing strategies.

In any context where changes are taking place continuously, calculus becomes indispensable. Specifically, the derivative provides a snapshot of how a cost or production rate changes at any given production level, hence allowing for precise, immediate insights.
derivatives
Derivatives in calculus give us the tools to determine the rate of change of a function. In the realm of economic cost analysis, the derivative of a cost function, referred to as the marginal cost, enables firms to understand the cost implications of producing one additional unit.

The marginal cost is essentially the calculus derivative of a cost function, represented as \( C'(q) \), and defines how cost changes with respect to a change in quantity. Calculating a derivative is about finding how a minor change in the input (quantity) influences the output (cost). This provides immediate insights into efficiency and cost-saving opportunities, helping decision-makers optimize production and minimize unnecessary expenses.

To estimate the derivative numerically from data, we consider finite differences, where we analyze the change in cost over the corresponding change in quantity. For example, when estimating the marginal cost at a midpoint like \( q=2000 \), we evaluate data intervals that bracket 2000 tons, such as \([1500, 2000]\) and \([2000, 2500]\). By averaging these computed differences, a reliable marginal cost estimation can be achieved.
cost analysis
Cost analysis involves a detailed examination of all expenses involved in a business process or operation, helping organizations identify areas of financial improvement. A key component of cost analysis is the concept of marginal cost—understanding the cost to produce one more unit of a product.

This analysis allows businesses to pinpoint cost-saving opportunities and improve production efficiencies. By understanding marginal cost, firms can make data-driven decisions on whether to increase or decrease production. For instance, if the marginal cost is low, it might suggest scaling up production is economical. On the contrary, a high marginal cost might indicate inefficiencies or a need for different management strategies.

Furthermore, by identifying the intervals with the smallest marginal cost, firms can determine optimal production levels, thus reducing overall expenditure. Such insights contribute to strategic planning and budgeting, enabling firms to allocate resources more effectively and sustain profitability in competitive markets.

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Most popular questions from this chapter

A company's cost of producing \(q\) liters of a chemical is \(C(q)\) dollars; this quantity can be sold for \(R(q)\) dollars. Suppose \(C(2000)=5930\) and \(R(2000)=7780\) (a) What is the profit at a production level of \(2000 ?\) (b) If \(M C(2000)=2.1\) and \(M R(2000)=2.5,\) what is the approximate change in profit if \(q\) is increased from 2000 to \(2001 ?\) Should the company increase or decrease production from \(q=2000 ?\) (c) If \(M C(2000)=4.77\) and \(M R(2000)=4.32\) should the company increase or decrease production from \(q=2000 ?\)

The size, \(S,\) of a tumor (in cubic millimeters) is given by \(S=2^{t},\) where \(t\) is the number of months since the tumor was discovered. Give units with your answers. (a) What is the total change in the size of the tumor during the first six months? (b) What is the average rate of change in the size of the tumor during the first six months? (c) Estimate the rate at which the tumor is growing at \(t=6 .\) (Use smaller and smaller intervals.)

When you breathe, a muscle (called the diaphragm) reduces the pressure around your lungs and they expand to fill with air. The table shows the volume of a lung as a function of the reduction in pressure from the diaphragm. Pulmonologists (lung doctors) define the compliance of the lung as the derivative of this function. 10 (a) What are the units of compliance? (b) Estimate the maximum compliance of the lung. (c) Explain why the compliance gets small when the lung is nearly full (around 1 liter). $$\begin{array}{|c|c|} \hline \begin{array}{c} \text { Pressure reduction } \\ \text { (cm of water) } \end{array} & \begin{array}{c} \text { Volume } \\ \text { (iters) } \end{array} \\ \hline 0 & 0.20 \\ \hline 5 & 0.29 \\ \hline 10 & 0.49 \\ \hline 15 & 0.70 \\ \hline 20 & 0.86 \\ \hline 25 & 0.95 \\ \hline 30 & 1.00 \\ \hline \end{array}$$

The average weight, \(W,\) in pounds, of an adult is a function, \(W=f(c),\) of the average number of Calories per day, \(c,\) consumed. (a) Interpret the statements \(f(1800)=155\) and \(f^{\prime}(2000)=0\) in terms of diet and weight. (b) What are the units of \(f^{\prime}(c)=d W / d c ?\)

For the function \(f(x)=3^{x},\) estimate \(f^{\prime}(1) .\) From the graph of \(f(x),\) would you expect your estimate to be greater than or less than the true value of \(f^{\prime}(1) ?\)

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