/*! 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 9 The yearly inventory expense \(E... [FREE SOLUTION] | 91Ó°ÊÓ

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The yearly inventory expense \(E\), in dollars, of a car dealer is a function of the number \(Q\) of automobiles ordered at a time from the manufacturer. A dealer who orders only a few automobiles at a time will have the expense of placing several orders, whereas if the order sizes are large, then the dealer will have a large inventory of unsold automobiles. For one dealer the formula is $$ E=\frac{425 Q^{2}+8000}{Q}, $$ so \(E\) is a rational function of \(Q\). a. Make a graph of \(E\) versus \(Q\) covering order sizes up to \(10 .\) b. Explain in practical terms the behavior of the graph near the pole at \(Q=0\). (Hint: Keep in mind that there is a fixed cost of processing each order, regardless of the size of the order.)

Short Answer

Expert verified
As \( Q \to 0^+ \), \( E \to \infty \), reflecting high costs from frequent small orders. For larger \( Q \), costs stabilize, balancing order and holding expenses.

Step by step solution

01

Understand the Function

The given function for inventory expense is \( E = \frac{425 Q^{2} + 8000}{Q} \). It is a rational function, where \( E \) is expressed in terms of \( Q \), the number of automobiles ordered.
02

Simplify the Equation

Rewrite the function: \( E = 425Q + \frac{8000}{Q} \). This form highlights that \( E \) is a combination of linear and reciprocal terms in \( Q \).
03

Determine Domain

Since \( Q \) represents the number of automobiles, it must be positive. Therefore, \( Q > 0 \). Note the graph will have a vertical asymptote at \( Q = 0 \), since division by zero is undefined.
04

Analyze Behavior Near the Pole

As \( Q \) approaches 0 from the right, the term \( \frac{8000}{Q} \) becomes very large, and therefore \( E \) increases. This reflects the high cost when ordering very few automobiles.
05

Graph the Function \( E(Q) \) for \( 1 \leq Q \leq 10 \)

Calculate \( E(Q) \) for integers from 1 to 10, plot these points, and sketch the curve. For example, if \( Q = 1 \), \( E = 425 \cdot 1 + \frac{8000}{1} = 8425 \). Compute similarly for other \( Q \) values.
06

Explain Graph Behavior at \( Q = 0 \)

Near \( Q = 0 \), costs are high due to frequent ordering despite a large fixed cost per order. As \( Q \) increases, the cost tends to stabilize, reflecting the trade-off between ordering costs and holding costs.

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Key Concepts

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

Inventory Expense
Inventory expense is a crucial concept in managing costs for businesses, especially for those dealing with physical products like cars. It represents the cost related to handling, storing, and maintaining a stock of goods. For a car dealer, this expense depends on the strategy of ordering vehicles from the manufacturer.

When a dealer orders a small number of cars, the inventory expense is mainly driven by the fixed costs associated with processing each order. This means placing multiple smaller orders can lead to higher overall costs because each order incurs a distinct processing fee. On the flip side, ordering a large quantity of cars at once can lower the frequency of orders, but it increases the storage costs due to having more unsold inventory.

The given formula, \(E = \frac{425 Q^{2} + 8000}{Q}\), captures this relationship as a rational function. This equation indicates that the expense grows with both the number of cars ordered and the handling of these items while balancing the holding costs.
Asymptote
Asymptotes are lines that a graph approaches but never actually touches or crosses. Understanding asymptotes helps in predicting the behavior of rational functions at extreme values.

For the inventory expense function \(E = \frac{425 Q^{2} + 8000}{Q}\), there is a vertical asymptote at \(Q = 0\). This asymptote occurs because the expression involves division by \(Q\), and division by zero is undefined. Consequently, as \(Q\) becomes very small (approaching zero but staying positive), the value of \(E\) increases drastically, which indicates a very high cost.

This infinite increase reflects the scenario where ordering becomes increasingly inefficient as fewer items are ordered at a time, while still incurring high fixed costs. In practical terms, the dealer faces escalating costs when attempting to manage their inventory by frequently ordering a minimal amount of cars.
Domain of a Function
The domain of a function refers to all the possible input values that will result in a valid output. In simple terms, it indicates the "legal" values for \(Q\) in the function \(E = \frac{425 Q^{2} + 8000}{Q}\).

Here, \(Q\) represents the number of cars ordered, naturally limiting \(Q\) to positive integers, because you cannot order a negative number of cars. Therefore, the domain is all positive numbers, or mathematically, \(Q > 0\).

Choosing a domain ensures that the calculations within the function do not involve any illegal operations, like division by zero, which would occur at \(Q = 0\). Additionally, considering the context of ordering vehicles, it makes practical sense to restrict \(Q\) to positive integers, as orders in fractions or negative values aren't feasible in real-life scenarios.

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Most popular questions from this chapter

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