/*! 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 27 Assume that the demand equation ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 27

Assume that the demand equation for tuna in a small coastal town is $$p q^{1.5}=50,000$$ where \(q\) is the number of pounds of tuna that can be sold in one month at the price of \(p\) dollars per pound. The town's fishery finds that the demand for tuna is currently 900 pounds per month and is increasing at a rate of 100 pounds per month each month. How fast is the price changing?

Short Answer

Expert verified
The price (\(p\)) is decreasing at a rate of approximately $0.0219 per pound per month.

Step by step solution

01

Given parameters

: We are given the following: - The demand equation for tuna: \(pq^{1.5} = 50,000\) - Current demand for tuna (\(q\)): 900 pounds per month - Rate of increase in demand (\(\frac{dq}{dt}\)): 100 pounds per month
02

Compute the current price p

: We need to find the current price (\(p\)) for the given demand (\(q = 900\)). To do this, we can plug the value of \(q\) into the demand equation and solve for \(p\): \(p(900)^{1.5} = 50,000\) Divide both sides of the equation by \((900)^{1.5}\): \(p = \frac{50,000}{(900)^{1.5}}\) Now we calculate the value of p: \(p \approx 1.95\) The current price (\(p\)) is approximately $1.95 per pound.
03

Differentiate the demand equation

: To find the rate of change of the price (\(\frac{dp}{dt}\)), we need to differentiate the demand equation with respect to time (\(t\)). Using the chain rule, we can rewrite the demand equation as: \((pq^{1.5})' = (pdq^{1.5/dt}) + (q^{1.5}dp/dt) = 0\)
04

Plug in the given values and solve for dp/dt

: Now we will plug in the given values of \(q\), \(p\), and \(\frac{dq}{dt}\), and solve for \(\frac{dp}{dt}\): \(p(\frac{dq}{dt}q^{0.5}) + (q^{1.5}\frac{dp}{dt}) = 0\) Using the values: \(p \approx 1.95\), \(q = 900\), and \(\frac{dq}{dt} = 100\): \((1.95)(100)(900)^{0.5} + (900)^{1.5}\frac{dp}{dt} = 0\) Now, solve for \(\frac{dp}{dt}\): \((900)^{1.5} \frac{dp}{dt} = -1.95(100)(900)^{0.5}\) Finally, calculate \(\frac{dp}{dt}\): \(\frac{dp}{dt} \approx -0.0219\) Thus, the price (\(p\)) is decreasing at a rate of approximately $0.0219 per pound 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 91影视!

Key Concepts

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

Demand Equation
The demand equation is used to relate the price of a product to the quantity demanded. In our case, the demand equation for tuna in the coastal town is expressed as \(p q^{1.5} = 50,000\). This tells us the relationship between the price \(p\) and the amount of tuna \(q\) that can be sold. It essentially says that as either the price or demand changes, they work together to balance the equation.
To fully understand demand equations, remember that:
  • \(p\) represents the price per unit (here, the price per pound of tuna).
  • \(q\) denotes the quantity demanded in a specific time period.
In our problem, knowing this demand equation helped us establish and compute the current conditions, such as finding the value of \(p\) when \(q\) is at 900 pounds.
Rate of Change
The "rate of change" refers to how fast a variable changes with respect to time. For this exercise, we are interested in how quickly the price of tuna \(p\) is changing given that the demand \(q\) is increasing. The rate at which demand is increasing is given as \(\frac{dq}{dt} = 100\) pounds per month.
In general, the rate of change can be linked to the derivative of a function: if we have a function \(f(t)\), then \(\frac{df}{dt}\) represents its rate of change over time.
So, when we say the rate of change of the price, \(\frac{dp}{dt}\), it means we are looking for how quickly the price changes as time moves forward, considering that the underlying demand changes.
Differentiation with Respect to Time
Differentiating with respect to time involves using calculus to understand how variables change over time. In this exercise, we applied differentiation to the demand equation in order to find how the price, \(p\), changes as demand, \(q\), increases. When differentiating \( \( p q^{1.5} \) \), we used the chain rule from calculus, which is essential when dealing with functions dependent on time.
This involves treating each term involving time as a function of time and applying the rule which states:
  • The derivative of a product is the derivative of the first term times the second, plus the first term times the derivative of the second.
By applying this rule:
  • Differentiate \( p \), the variable we're interested in.
  • Also differentiate \( q^{1.5} \), since \( q \) changes over time at rate \( \frac{dq}{dt} \).
Finally, solving the resulting equation by plugging in the given values helped us find \( \frac{dp}{dt} \), the desired rate at which the price changes.

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

Daily oil production in Mexico and daily U.S. oil imports from Mexico during \(2005-2009\) can be approximated by $$\begin{array}{ll}P(t)=3.9-0.10 t \text { million barrels } & (5 \leq t \leq 9) \\ I(t)=2.1-0.11 t \text { million barrels } & (5 \leq t \leq 9) \end{array}$$ where \(t\) is time in years since the start of \(2000 .^{41}\) Graph the function \(I(t) / P(t)\) and its derivative. Is the graph of \(I(t) / P(t)\) concave up or concave down? The concavity of \(I(t) / P(t)\) tells you that: (A) The percentage of oil produced in Mexico that was exported to the United States was decreasing. (B) The percentage of oil produced in Mexico that was not exported to the United States was increasing. (C) The percentage of oil produced in Mexico that was exported to the United States was decreasing at a slower rate. (D) The percentage of oil produced in Mexico that was exported to the United States was decreasing at a faster rate.

The average cost function for the weekly manufacture of portable CD players is given by \(\bar{C}(x)=150,000 x^{-1}+20+0.0001 x\) dollars per player, where \(x\) is the number of CD players manufactured that week. Weekly production is currently 3,000 players and is increasing at a rate of 100 players per week. What is happening to the average cost? HINT [See Example 3.]

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 ) \(^{42}\)

Sketch the graph of the given function, indicating (a) \(x\) - and \(y\) -intercepts, (b) extrema, (c) points of inflection, \((d)\) behavior near points where the function is not defined, and (e) behavior at infinity. Where indicated, technology should be used to approximate the intercepts, coordinates of extrema, and/or points of inflection to one decimal place. Check your sketch using technology. \(f(x)=x+\frac{1}{x}\)

The Physics Club sells \(E=m c^{2}\) T-shirts at the local flea market. Unfortunately, the club's previous administration has been losing money for years, so you decide to do an analysis of the sales. A quadratic regression based on old sales data reveals the following demand equation for the T-shirts: $$q=-2 p^{2}+33 p . \quad(9 \leq p \leq 15)$$ Here, \(p\) is the price the club charges per T-shirt, and \(q\) is the number it can sell each day at the flea market. a. Obtain a formula for the price elasticity of demand for \(E=m c^{2}\) T-shirts. b. Compute the elasticity of demand if the price is set at \(\$ 10\) per shirt. Interpret the result. c. How much should the Physics Club charge for the T-shirts in order to obtain the maximum daily revenue? What will this revenue be?
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation

Mechanics Maths

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.