/*! 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 23 Your company manufactures automo... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Your company manufactures automobile alternators, and production is partially automated through the use of robots. In order to meet production deadlines, your company calculates that the numbers of laborers and robots must satisfy the constraint $$x y=10,000$$ where \(x\) is the number of laborers and \(y\) is the number of robots. Your company currently uses 400 robots and is increasing robot deployment at a rate of 16 per month. How fast is it laying off laborers? HINT [See Example 4.

Short Answer

Expert verified
The company is laying off laborers at a rate of 1 per month.

Step by step solution

01

Differentiate both sides with respect to time

Implicitly differentiate the given constraint, x*y = 10,000, with respect to time (t). We'll use the chain rule for this: \(\frac{dx}{dt}y + x\frac{dy}{dt} = 0\) In this step, we implicitly differentiated both sides with respect to time. We are now left with a new equation involving the rates of change of x and y.
02

Plug in the given values

Now plug in the values provided: number of current robots (y = 400), the increase in robots (dy/dt = 16), and calculate the number of laborers using the constraint (x = 10,000/y): \(\frac{dx}{dt}(400) + x(16) = 0\) We can find the value of x by plugging in the value of y into the constraint equation: \(x(400) = 10,000\) \(x = \frac{10,000}{400} = 25\) So, the number of laborers is 25.
03

Solve for dx/dt

Plug in the value of x into the differentiated equation and solve for dx/dt: \(\frac{dx}{dt}(400) + 25(16) = 0\) \(400\frac{dx}{dt} = -25(16)\) \(\frac{dx}{dt} = \frac{-25(16)}{400}\) \(\frac{dx}{dt} = -1\) So, the company is laying off laborers at a rate of 1 per month.

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 91Ó°ÊÓ!

Key Concepts

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

Implicit Differentiation
Implicit differentiation is a powerful technique used in calculus when dealing with equations in which the variables cannot be separated easily. This occurs often in related rates problems when defining a relationship between two or more variables that are all changing with respect to time.

In our exercise, the relationship between the number of laborers (\(x\)) and robots (\(y\) ) is given as an implicit function \(xy = 10,000\textbackslash\textbackslash\textbackslash\textbackslash\textbackslash\textbackslash\textbackslash\textbackslash\textbackslash\textbackslash\textbackslash\textbackslash\textbackslash\textbackslash\textbackslash\textbackslash\textbackslash\textbackslash\). Instead of solving for \(y\) or \(x\) directly, we differentiate both sides with respect to time (\(t\)), acknowledging that \(x\) and \(y\) are functions of \(t\) due to their rates of change. By applying this technique, we can find out how the rate of change of one variable is affecting the other.
Product Rule
The product rule is an essential differentiation rule used when taking the derivative of products of two functions. In related rates problems such as ours, the product rule becomes necessary when differentiating an equation where one variable (laborers) is multiplied by another (robots).

The equation \(xy = 10,000\) involves the product of \(x\) and \(y\). When differentiating with respect to time, we apply the product rule which states that to differentiate \(xy\), we take the derivative of \(x\) times \(y\), plus \(x\) times the derivative of \(y\), or formally: \(\frac{d}{dt}(xy) = \(x\) \(\frac{dy}{dt}\) + \(y\) \(\frac{dx}{dt}\).\) By correctly applying the product rule, we gain insight into how changes in one variable impact the other, which is vital for finding our solution.
Differentiation With Respect to Time
Differentiation with respect to time is the process of finding the rate at which a quantity changes over time. In calculus, this concept is frequently used to connect spatial changes with temporal changes in related rates problems.

In practice, we denote differentiation with respect to time by using the notation \(\frac{d}{dt}\), where \(t\) stands for time. For instance, when we discuss how fast the number of robots (\(y\)) is increasing, we use \(\frac{dy}{dt}\) to represent this rate. Similarly, \(\frac{dx}{dt}\) represents the rate of change of the number of laborers. Balancing the rates of these changes according to the given constraint allows us to solve for the unknown rates effectively.
Real-World Application of Calculus
Calculus isn't just abstract theory; it's a practical tool used to solve real-world problems. In the given exercise, we explored a real-world application of calculus in a manufacturing context. By using related rates calculus, we evaluated how the increase in robots affects the number of laborers needed to meet production targets.

This problem is a small reflection of how calculus helps businesses make critical decisions for efficiency and cost savings. Everyday applications include optimizing production, calculating rates of change in finance, and even predicting the spread of diseases in epidemiology. The ability to relate the rate of change in one variable to that of another—like laborers to robots—can provide valuable insights and inform strategy in various industries.

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 production formula for a student's performance on a difficult English examination is given by $$g=4 h x-0.2 h^{2}-10 x^{2}$$ where \(g\) is the grade the student can expect to obtain, \(h\) is the number of hours of study for the examination, and \(x\) is the student's grade point average. The instructor finds that students' grade point averages have remained constant at \(3.0\) over the years, and that students currently spend an average of 15 hours studying for the examination. However, scores on the examination are dropping at a rate of 10 points per year. At what rate is the average study time decreasing?

Repeat Exercise 81 using instead the models for 2000-2004 shown below: $$\begin{array}{ll}P(t)=3.0+0.13 t \text { million barrels } & (0 \leq t \leq 4) \\\I(t)=1.4+0.06 t \text { million barrels } & (0 \leq t \leq 4)\end{array} $$( \(t\) is time in years since the start of 2000 ) \(^{42}\)

The percentage of U.S.-issued mortgages that were subprime can be approximated by $$A(t)=\frac{15.0}{1+8.6(1.8)^{-t}} \text { percent } \quad(0 \leq t \leq 8)$$ \(t\) years after the start of \(2000^{52}\) Graph the derivative \(A^{\prime}(t)\) of \(A(t)\) using an extended domain of \(0 \leq t \leq 15\). Determine the approximate coordinates of the maximum and determine the behavior of \(A^{\prime}(t)\) at infinity. What do the answers tell you?

The demand for personal computers in the home goes up with household income. For a given community, we can approximate the average number of computers in a home as $$q=0.3454 \ln x-3.047 \quad 10,000 \leq x \leq 125,000$$ where \(x\) is mean household income. \({ }^{60}\) Your community has a mean income of \(\$ 30,000\), increasing at a rate of \(\$ 2,000\) per year. How many computers per household are there, and how fast is the number of computers in a home increasing? (Round your answer to four decimal places.)

On a recent exam, you were given a related rates problem based on an algebraic equation relating two variables \(x\) and \(y\). Your friend told you that the correct relationship between \(d x / d t\) and \(d y / d t\) was given by $$\left(\frac{d x}{d t}\right)=\left(\frac{d y}{d t}\right)^{2}$$ Could he be correct?

See all solutions

Recommended explanations on Math Textbooks

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.