/*! 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 24 The production function \(f(x)\)... [FREE SOLUTION] | 91Ó°ÊÓ

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The production function \(f(x)\) gives the number of units of an item that a manufacturing company can produce from \(x\) units of raw material. The company buys the raw material at price \(w\) dollars per unit and sells all it produces at a price of \(p\) dollars per unit. The quantity of raw material that maximizes profit is denoted by \(x^{*}\) (a) Do you expect the derivative \(f^{\prime}(x)\) to be positive or negative? Justify your answer. (b) Explain why the formula \(\pi(x)=p f(x)-w x\) gives the profit \(\pi(x)\) that the company earns as a function of the quantity \(x\) of raw materials that it uses. (c) Evaluate \(f^{\prime}\left(x^{*}\right)\) (d) Assuming it is nonzero, is \(f^{\prime \prime}\left(x^{*}\right)\) positive or negative? (e) If the supplier of the raw materials is likely to change the price \(w,\) then it is appropriate to treat \(x^{*}\) as a function of \(w .\) Find a formula for the derivative \(d x^{*} / d w\) and decide whether it is positive or negative. (f) If the price \(w\) goes up, should the manufacturing company buy more or less of the raw material?

Short Answer

Expert verified
(a) Positive. (b) \(\pi(x)\) is revenue minus cost. (c) \(f'(x^*) = \frac{w}{p}\). (d) Negative. (e) \(dx^*/dw < 0\), negative. (f) Buy less raw material.

Step by step solution

01

Analyze the Derivative of the Production Function

For part (a), we consider whether the derivative of the production function, \( f'(x) \), is positive or negative. The production function \( f(x) \) denotes the output produced from \( x \) units of raw material. Typically, an increase in raw material \( x \) should lead to an increase in output (at least up to a point), making \( f'(x) > 0 \). Thus, we expect \( f'(x) \) to be positive because more input generally leads to more output.
02

Establish the Profit Function

For part (b), we analyze the profit formula \( \pi(x) = p f(x) - wx \). \( p f(x) \) represents the total revenue, where \( p \) is the price per unit sales and \( f(x) \) is the production. Therefore, \( pf(x) \) is the dollar amount earned from selling the units produced. \( wx \) represents the total cost of the raw material. Thus, \( \pi(x) \) captures the profit, being the difference between total revenue and total cost.
03

Evaluate Derivative at Maximum Profit

In part (c), we determine that the optimal quantity \( x^* \) maximizes profit \( \pi(x) \). At \( x^* \), the derivative of the profit with respect to \( x \), \( \pi'(x) = p f'(x) - w \), must be zero (first-order condition for a maximum). Thus, \( p f'(x^*) = w \), giving \( f'(x^*) = \frac{w}{p} \).
04

Second Derivative Test

In part (d), we need to identify whether \( f''(x^*) \) is positive or negative given that it is nonzero. The second derivative \( \pi''(x) = p f''(x) \) indicates the concavity of the profit function. For maximizing profit, \( \pi''(x) < 0 \), implying \( f''(x^*) < 0 \). Thus, \( f''(x^*) \) is negative.
05

Derive Expression for Change in Optimal Raw Material

In part (e), if the price \( w \) changes, we treat \( x^* \) as a function of \( w \). Using implicit differentiation on \( pf'(x^*) = w \), with respect to \( w \), gives \( p f''(x^*) \frac{dx^*}{dw} = 1 \). Solving gives \( \frac{dx^*}{dw} = \frac{1}{p f''(x^*)} \). Since \( f''(x^*) < 0 \), \( \frac{dx^*}{dw} < 0 \). Thus, \( x^* \) decreases if \( w \) increases.
06

Effect of Increase in Raw Material Price

For part (f), since \( \frac{dx^*}{dw} < 0 \), an increase in \( w \) should lead to a decrease in \( x^* \). Therefore, the manufacturing company should buy less raw material if the price \( w \) increases.

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Key Concepts

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

Production Function
A production function is central in economics as it describes the relationship between input and output. In this context, the production function, denoted as \( f(x) \), provides the number of units a company can produce from \( x \) units of raw material. It essentially measures the efficiency and capacity of production. Here's why it matters:
  • It helps in understanding how varying the inputs will affect the output.
  • It aids in estimating the cost of production by determining resource needs.
  • It allows businesses to plan production quantities and allocate resources efficiently.
In most scenarios, increasing the input material \( x \) should lead to a rise in output up to a certain limit, indicating that the derivative \( f'(x) \) is typically positive. This positive derivative shows increasing returns to input, often assumed in production functions to model realistic manufacturing processes.
Derivative
The derivative of the production function, \( f'(x) \), is a crucial analytical tool to understand the rate of change of output with respect to input. It signifies how much additional output can be achieved by using one more unit of the input material:
  • If \( f'(x) > 0 \), it means that an increase in input leads to more output, which is usually the case, reflecting efficiency in resource usage.
  • The value of \( f'(x) \) at \( x^* \), the maximum profit point, balances revenue per additional unit of product and cost per unit of input, making \( f'(x^*) = \frac{w}{p} \).
Understanding the derivative is essential for making production decisions, as it helps indicate the responsiveness of the production output to changes in input.
Cost Analysis
Cost analysis focuses on understanding the expenses involved in the production process. In this exercise, the cost incurred by the company is \( wx \), where \( w \) is the cost per unit of raw material. Here's how it plays a role in profit calculation:
  • Determines how input costs accumulate as production scales up.
  • Influences pricing decisions and long-term planning to ensure profitability.
  • Forms a critical component of the profit equation \( \pi(x) = p f(x) - wx \), where the goal is to maximize the difference between revenue and costs.
In cost analysis, companies learn how variations in input cost impact total expenses and subsequently adjust their input levels (\( x^* \) in this context) to align with cost and revenue changes.
Revenue Function
The revenue function represents the total income from sales of the product, expressed as \( p f(x) \), where \( p \) is the selling price per unit. This function is crucial in assessing business performance and strategies:
  • It identifies how production levels influence overall revenue.
  • Helps to forecast financial outcomes based on production and market conditions.
  • Aligns production output with market demand to optimize sales.
In this setup, businesses strive to manipulate their production functions to maximize the revenue function, ensuring that sales optimize the resources invested. Comprehension of this function is necessary for strategic pricing and production plans to maximize profitability.

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Most popular questions from this chapter

On February \(16,2007,\) paraglider Eva Wisnierska \(^{11}\) was caught in a freak thunderstorm over Australia and carried upward at a speed of about \(3000 \mathrm{ft} / \mathrm{min.}\). Table 4.4 gives the temperature at various heights. Approximately how fast (in \(^{\circ} \mathrm{F} /\) per minute) was her temperature decreasing when she was at 4000 feet? $$\begin{array}{l|c|c|c|c|c|c|c|c} \hline y \text { (thousand ft) } & 2 & 4 & 6 & 8 & 10 & 12 & 14 & 16 \\ \hline H\left(^{\circ} \mathrm{F}\right) & 60 & 52 & 38 & 31 & 23 & 16 & 9 & 2 \\\ \hline \end{array}$$

Give an example of: A function with local maxima and minima at an infinite number of points.

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For positive constants \(k\) and \(g\), the velocity, \(v\), of a particle of mass \(m\) at time \(t\) is given by $$v=\frac{m g}{k}\left(1-e^{-k t / m}\right)$$ At what rate is the velocity changing at time 0? At \(t=1 ?\) What do your answers tell you about the motion?

Explain what is wrong with the statement. Every function of the form \(f(x)=x^{2}+b x+c,\) where \(b\) and \(c\) are constants, has two zeros.

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