/*! 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 74 The cost \(C\) for ordering and ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 74

The cost \(C\) for ordering and storing \(x\) units is \(C=2 x+300,000 / x .\) What order size will produce a minimum cost?

Short Answer

Expert verified
The order size that will result in a minimum cost is \(\sqrt{150,000} units\).

Step by step solution

01

Find the derivative of the cost function

First, find the derivative of the cost function. The derivative of \(C(x) = 2x + \frac{300,000}{x}\) is \(C'(x) = 2 - \frac{300,000}{x^2}\), which results from applying the Power Rule to the first term and the Quotient rule to the second term
02

Determine the critical point

Next, find the value of \(x\) that will yield a zero derivative by setting \(C'(x)\) equal to zero. This gives \(2 - \frac{300,000}{x^2} = 0\). Solving for \(x\), we obtain \(x = \sqrt{\frac{300,000}{2}} = \sqrt{150,000}\)
03

Ensure that the critical point yields a minimum value

Now, we need to confirm if \(x = \sqrt{150,000}\) indeed results to a minimum cost. This can be done by either calculating the second derivative or reasoning based on the nature of the cost function \(C(x) = 2x + \frac{300,000}{x}\) which is always positive. Since ordering less than or more than \(\sqrt{150,000}\) units will result in added costs due to under-utilising the fixed cost or due to increased marginal ordering cost, it can be reasoned that \(x = \sqrt{150,000}\) is indeed a minimum point.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A retailer has determined the cost \(C\) for ordering and storing \(x\) units of a product to be modeled by \(C=3 x+\frac{20,000}{x}, 0

In Exercises, use a graphing utility to find graphically the absolute extrema of the function on the closed interval. $$ f(x)=3.2 x^{5}+5 x^{3}-3.5 x, \quad[0,1] $$

In Exercises, identify the point of diminishing returns for the input output function. For each function, \(R\) is the revenue and \(x\) is the amount spent on advertising. Use a graphing utility to verify your results. $$ R=\frac{1}{50,000}\left(600 x^{2}-x^{3}\right), \quad 0 \leq x \leq 400 $$

In Exercises, use a graphing utility to graph the function. Then find all relative extrema of the function. $$ f(t)=(t-1)^{1 / 3} $$

In Exercises, you are given \(f^{\prime}\). Find the intervals on which (a) \(f^{\prime}(x)\) is increasing or decreasing and (b) the graph of \(f\) is concave upward or concave downward. (c) Find the relative extrema and inflection points of \(f\). (d) Then sketch a graph of \(f\). $$ f(x)=3 x^{2}-2 $$
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Calculus

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.