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The average cost (in dollars) incurred by Lincoln Records each week in pressing \(x\) compact discs is given by $$ \bar{C}(x)=-0.0001 x+2+\frac{2000}{x} \quad(0

Short Answer

Expert verified
The derivative of the average cost function \(\bar{C}(x) = -0.0001 x + 2 + \frac{2000}{x}\) is \(\bar{C}'(x) = -0.0001 - \frac{2000}{x^2}\). Since both terms in \(\bar{C}'(x)\) are negative for \(x > 0\), this means that \(\bar{C}'(x) < 0\) over the interval (0, 6000), indicating that the average cost function \(\bar{C}(x)\) is always decreasing in this interval.

Step by step solution

01

Find the first derivative of \(\bar{C}(x)\)

The given function is \(\bar{C}(x)=-0.0001 x+2+\frac{2000}{x}\). To find its derivative, differentiating it with respect to \(x\). \[ \begin{aligned} \dfrac{d}{dx}\left(-0.0001 x+2+\frac{2000}{x}\right)=& \dfrac{d}{dx}\left(-0.0001 x\right)+\dfrac{d}{dx}\left(2\right)+\dfrac{d}{dx}\left(\frac{2000}{x}\right) \\ =& -0.0001\cdot \dfrac{d}{dx}\left(x\right)+0+\dfrac{d}{dx}\left(x^{-1}\right) \cdot 2000 \\ =& -0.0001+\left(-1\right) \cdot x^{-2} \cdot 2000 \\ =& -0.0001-\frac{2000}{x^2} \end{aligned} \] So, the first derivative is \(\bar{C}'(x) = -0.0001-\frac{2000}{x^2}\).
02

Determine the sign of \(\bar{C}'(x)\) over the interval \((0, 6000)\)

To show that \(\bar{C}(x)\) is always decreasing over the interval \((0, 6000)\), we must check if \(\bar{C}'(x)\) is negative over this interval. Since both the terms in \(\bar{C}'(x)\), -0.0001 and -\(\frac{2000}{x^2}\), are negative and \(x>0\), their sum will also be negative over the entire interval. As \(\bar{C}'(x) < 0\) over the interval \((0, 6000)\), we can conclude that the average cost function \(\bar{C}(x)\) is always decreasing over this interval.

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Key Concepts

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

Derivative
In calculus, the derivative of a function measures how the function's output value changes as its input changes. Think of it like the function's speedometer, telling you how fast or slow the function climbs or descends. For the cost function \( \bar{C}(x) = -0.0001 x + 2 + \frac{2000}{x} \), we find its first derivative to understand the behavior of the cost with respect to the number of compact discs produced.

To find the derivative \( \bar{C}'(x) \), we apply basic differentiation rules: the power rule and the constant multiple rule. The power rule helps with terms like \( \frac{2000}{x} \), which simplifies to \( 2000x^{-1} \). Differentiating \( 2000x^{-1} \) yields \(-2000x^{-2}\), showcasing how the presence of \(x^2\) in the denominator affects the rate of change. The term \(-0.0001 x\) becomes \(-0.0001\) upon differentiation, while constants like \(2\) vanish.

The resulting derivative, \( \bar{C}'(x) = -0.0001 - \frac{2000}{x^2} \), gives insights into the smoothness or steepness at each point. This mathematical tool is essential for analyzing how sensitive a function is to change.
Decreasing Function
A function is said to be decreasing when its overall value diminishes as its input increases. In simpler terms, if you imagine walking downhill, each step forward corresponds to a lower elevation — that's the intuitive grasp of a decreasing function.

For \( \bar{C}(x) \) to be termed a decreasing function over the interval \((0, 6000)\), its derivative \( \bar{C}'(x) \) must consistently stay negative throughout this interval. In our solution, the calculated \( \bar{C}'(x) = -0.0001 - \frac{2000}{x^2} \) is composed of two negative parts for any positive \(x\). Both terms—\(-0.0001\) and \(-\frac{2000}{x^2}\)—are negative regardless of the \(x\) values within \( (0, 6000) \).

This combination ensures that \( \bar{C}'(x) \) remains negative, confirming that the function is truly decreasing. This characteristic indicates that producing more compact discs costs less per unit, a valuable insight for optimizing production.
Cost Function
In economics and business, understanding the cost function is vital for decision-making. A cost function, like \( \bar{C}(x) \), relates the cost of production to the number of items produced. It helps in calculating the average or total costs as production scales up or down and forms the basis for pricing and profitability analysis.

The given cost function is \( \bar{C}(x) = -0.0001 x + 2 + \frac{2000}{x} \). This particular form includes both a linear component \(-0.0001 x\) and a rational component \(\frac{2000}{x}\). The linear part suggests a minor consistent cost reduction per additional unit produced, while the rational term \(\frac{2000}{x}\) highlights a significant drop in cost as more units are created.

The practical aim of understanding this function is optimizing production levels to minimize average costs. By examining how this function changes, businesses like Lincoln Records make decisions to increase efficiency and profitability, adjusting their operations to stay competitive.

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Most popular questions from this chapter

In an autocatalytic chemical reaction, the product formed acts as a catalyst for the reaction. If \(Q\) is the amount of the original substrate present initially and \(x\) is the amount of catalyst formed, then the rate of change of the chemical reaction with respect to the amount of catalyst present in the reaction is $$ R(x)=k x(Q-x) \quad(0 \leq x \leq Q) $$ where \(k\) is a constant. Show that the rate of the chemical reaction is greatest at the point when exactly half of the original substrate has been transformed.

The average revenue is defined as the function $$ \bar{R}(x)=\frac{R(x)}{x} \quad(x>0) $$ Prove that if a revenue function \(R(x)\) is concave downward \(\left[R^{\prime \prime}(x)<0\right]\), then the level of sales that will result in the largest average revenue occurs when \(\bar{R}(x)=R^{\prime}(x)\).

A manufacturer of tennis rackets finds that the total cost \(C(x)\) (in dollars) of manufacturing \(x\) rackets/day is given by \(C(x)=400+4 x+0.0001 x^{2}\). Each racket can be sold at a price of \(p\) dollars, where \(p\) is related to \(x\) by the demand equation \(p=10-0.0004 x\). If all rackets that are manufactured can be sold, find the daily level of production that will yield a maximum profit for the manufacturer.

Find the absolute maximum value and the absolute minimum value, if any, of each function. $$ f(x)=x e^{-x^{2}} \text { on }[0,2] $$

Find the absolute maximum value and the absolute minimum value, if any, of each function. $$ f(x)=x-\ln x \text { on }\left[\frac{1}{2}, 3\right] $$

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