/*! 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 11 A factory owner buys a new machi... [FREE SOLUTION] | 91Ó°ÊÓ

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A factory owner buys a new machine for \(\$ 20,000 .\) After eight years, the machine has a salvage value of \(\$ 1000 .\) Find a formula for the value of the machine after t years, where \(0 \leq t \leq 8\)

Short Answer

Expert verified
The value of the machine after t years is \(V(t) = 20000 - 2375t\).

Step by step solution

01

Identify the Initial Value and Salvage Value

The initial value of the machine is the purchase price, which is $20,000. The salvage value after 8 years is $1,000. These will be used to determine the depreciation rate of the machine.
02

Calculate Annual Depreciation

Depreciation is the reduction in value over time. Calculate annual depreciation by subtracting the salvage value from the initial value and then dividing by the number of years.\[ \text{Annual Depreciation} = \frac{\text{Initial Value} - \text{Salvage Value}}{\text{Number of Years}} = \frac{20000 - 1000}{8} = \$2375 \]
03

Formulate the Depreciation Formula

Using the annual depreciation, we can state the formula for the value of the machine after t years as: \[ V(t) = \text{Initial Value} - \text{Annual Depreciation} \times t = 20000 - 2375 \times t \]
04

Conclude the Formula

Reiterate the formula for clarity: The value of the machine after \(t\) years is \(V(t) = 20000 - 2375t\), where \(0 \leq t \leq 8\). This linear equation shows that the machine's value decreases by $2375 for each year.

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Key Concepts

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

Linear Depreciation
Linear depreciation is a straightforward way to account for the loss in value of an asset over time. Essentially, it's a method where the asset loses the same amount of value each year until it reaches its salvage value. This consistent rate of decline makes calculations simple and predictable.
Understanding linear depreciation is crucial because it offers a consistent way of accounting for asset value loss, making financial planning and reporting clearer for businesses.
In the context of the factory's machine, the value of the machine decreases evenly over its useful life of 8 years. Each year, it depreciates by the same dollar amount, allowing the owner to know exactly how much value it loses annually.
Initial Value
The initial value is the starting point for calculating depreciation. It's the cost of acquiring the asset. Think of it as the baseline from which all depreciation calculations begin.
In our exercise, the machine's initial value is $20,000. This price reflects what the factory owner paid for the machine when it was brand-new and fully functional.
Understanding the initial value is important because it sets a clear reference point for how much value the asset is intended to lose over its lifespan. Knowing the initial value also helps in determining insurance and pricing for replacement or selling purposes.
Salvage Value
The salvage value is the expected value of an asset at the end of its useful life. It's the estimated resale or scrap value that the asset might fetch once it's no longer useful for intended operations.
For the machine in the exercise, the salvage value after 8 years is set at $1,000. This amount is what the factory owner anticipates to recover from the machine once it is considered obsolete.
Recognizing the salvage value is important because it impacts how depreciation is calculated and reported. Knowing it in advance provides a better financial outlook, ensuring that the asset's depreciation doesn't exceed its actual use-value.
Depreciation Formula
The depreciation formula is the expression used to determine the asset's value at any given point in its lifespan. It's essential for both accounting and financial reporting, providing clarity on the diminishing value of an asset.
The formula in our exercise is based on the principle of straight-line depreciation: \[ V(t) = ext{Initial Value} - ext{Annual Depreciation} imes t = 20000 - 2375 imes t \] This formula calculates how much the machine's value decreases each year. As such, by plugging any given year, t, into the formula, you can easily find out the current value of the machine.
Understanding this formula helps manage business finances by predicting the asset's future worth, assisting both in budgeting and in deciding when new investments are needed.

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Most popular questions from this chapter

(a) Graph the four parabolas \(y=x^{2}+2 k x+1\) corresponding to \(k=2,3,0.75,1.5 .\) Among the four parabolas, which one appears to have the vertex closest to the origin? (b) Let \(f(x)=x^{2}+2 k x+1 .\) Find a positive value for \(k\) so that the distance from the origin to the vertex of the parabola is as small as possible. Check that your answer is consistent with your observations in part (a).

A piece of wire 14 in. long is cut into two pieces. The first piece is bent into a circle, the second into a square. Express the combined total area of the circle and the square as a function of \(x,\) where \(x\) denotes the length of the wire that is used for the circle.

Let \(f(x)=\left(x^{5}+1\right) / x^{2}\) (a) Graph the function \(f\) using a viewing rectangle that extends from -4 to 4 in the \(x\) -direction and from -8 to 8 in the \(y\) -direction. (b) Add the graph of the curve \(y=x^{3}\) to your picture in part (a). Note that as \(|x|\) increases (that is, as \(x\) moves away from the origin), the graph of \(f\) looks more and more like the curve \(y=x^{3} .\) For additional perspective, first change the viewing rectangle so that \(y\) extends from -20 to \(20 .\) (Retain the \(x\) -settings for the moment.) Describe what you see. Next, adjust the viewing rectangle so that \(x\) extends from -10 to 10 and \(y\) extends from -100 to \(100 .\) Summarize your observations. (c) In the text we said that a line is an asymptote for a curve if the distance between the line and the curve approaches zero as we move further and further out along the curve. The work in part (b) illustrates that a curve can behave like an asymptote for another curve. In particular, part (b) illustrates that the distance between the curve \(y=x^{3}\) and the graph of the given function \(f\) approaches zero as we move further and further out along the graph of \(f .\) That is, the curve \(y=x^{3}\) is an "asymptote" for the graph of the given function \(f\). Complete the following two tables for a numerical perspective on this. In the tables, \(d\) denotes the vertical distance between the curve \(y=x^{3}\) and the graph of \(f:\) $$ d=\left|\frac{x^{5}+1}{x^{2}}-x^{3}\right| $$ $$\begin{array}{llllll} \hline x & 5 & 10 & 50 & 100 & 500 \\ \hline d & & & & \\ \hline & & & & \\ \hline x & -5 & -10 & -50 & -100 & -500 \\ \hline d & & & & \\ \hline \end{array}$$ (d) Parts (b) and (c) have provided both a graphical and a numerical perspective. For an algebraic perspective that ties together the previous results, verify the following identity, and then use it to explain why the results in parts (b) and (c) were inevitable: $$ \frac{x^{5}+1}{x^{2}}=x^{3}+\frac{1}{x^{2}} $$

Let \(A B C\) be a triangle with vertices \(A(-3,0), B(0,6)\) \(C(3,0) .\) Let \(P\) be a point on the line segment that joins \(B\) to the origin. Find the position of \(P\) that minimizes the sum of the distances between \(P\) and the vertices.

Sketch the graph of each rational function. Specify the intercepts and the asymptotes. (a) \(f(x)=(x-1)(x+2.75) /[(x+1)(x+3)]\) (b) \(g(x)=(x-1)(x+3.25) /[(x+1)(x+3)]\) [Compare the graphs you obtain in parts (a) and (b). Notice how a relatively small change in one of the constants can radically alter the graph.]

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