/*! 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 29 Demand for Tents The quantity de... [FREE SOLUTION] | 91Ó°ÊÓ

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Demand for Tents The quantity demanded of the Sportsman \(5 \times 7\) tents, \(x\), is related to the unit price, \(p\), by the function $$ p=f(x)=-0.1 x^{2}-x+40 $$ where \(p\) is measured in dollars and \(x\) is measured in units of a thousand. a. Find the average rate of change in the unit price of a tent if the quantity demanded is between 5000 and 5050 tents; between 5000 and 5010 tents. b. What is the rate of change of the unit price if the quantity demanded is 5000 ?

Short Answer

Expert verified
a. The average rate of change in the unit price of a tent between 5000 and 5050 tents is approximately -2.0 dollars per thousand tents, and between 5000 and 5010 tents, it is approximately -2.01 dollars per thousand tents. b. The rate of change of the unit price if the quantity demanded is 5000 tents is -1001 dollars per thousand tents.

Step by step solution

01

Calculate the average rate of change for 5000-5050 tents

To find the average rate of change, use the formula: Average rate of change = \(\frac{f(x_2) - f(x_1)}{x_2 - x_1}\) Here, \(x_1 = 5000\) and \(x_2 = 5050\). First, evaluate f(5000) and f(5050): \(f(5000) = -0.1(5000)^2 - 5000 + 40\) \(f(5050) = -0.1(5050)^2 - 5050 + 40\) Average rate of change = \(\frac{f(5050) - f(5000)}{5050 - 5000}\)
02

Calculate the average rate of change for 5000-5010 tents

Similar to step 1, we will find the average rate of change for 5000-5010 tents. Here, \(x_1 = 5000\) and \(x_2 = 5010\). First, evaluate f(5010): \(f(5010) = -0.1(5010)^2 - 5010 + 40\) Average rate of change = \(\frac{f(5010) - f(5000)}{5010 - 5000}\) b. What is the rate of change of the unit price if the quantity demanded is 5000?
03

Find the derivative of the demand function

The rate of change of the unit price with respect to quantity demanded can be found by taking the derivative of the function f(x) with respect to x: \(f'(x) = \frac{d}{dx}(-0.1x^2 - x + 40)\) Apply the power rule and the constant rule: \(f'(x) = -0.2x - 1\)
04

Evaluate the derivative at x=5000

To find the rate of change of the unit price when the quantity demanded is 5000 tents, plug x=5000 into the derivative: \(f'(5000) = -0.2(5000) - 1\) Now compute the value of f'(5000). This will give you the rate of change of the unit price if the quantity demanded is 5000 tents.

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Key Concepts

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

Average Rate of Change
The average rate of change helps to determine how one quantity changes in relation to another over a specific interval. In calculus, it mimics how fast something changes over a particular range. For the demand problem, you're finding how the unit price of tents changes as you increase the number of tents sold.

To calculate this, use the formula:
  • Average rate of change = \(\frac{f(x_2) - f(x_1)}{x_2 - x_1}\).
Here, \(x_1\) and \(x_2\) represent the starting and ending quantities demanded (in thousands). Plug these into the demand function to find \(f(x_1)\) and \(f(x_2)\). This gives you the respective prices at these points.

By finding the difference in prices and dividing it by the difference in quantities, you get how much the price increased or decreased per unit sold between those quantities.
Derivative
Derivatives represent the rate at which a function is changing at any given point. In the context of your demand function, the derivative tells you how the price changes as the quantity of tents sold changes. It's the slope of the tangent line at any point on the demand curve and provides immediate information about the rate of change.

To find the derivative, differentiate the function \(f(x) = -0.1 x^{2} - x + 40\) with respect to \(x\). Apply differentiation rules to gain:\(f'(x) = -0.2x - 1\). This new function \(f'(x)\) gives the rate of change of price concerning the quantity at any point \(x\).

Evaluating this derivative at a specific point, like \(x = 5000\), answers questions on how the price is precisely changing with the sale of 5000 tents.
Demand Function
A demand function relates the quantity of an item demanded to its price. Understanding it can help businesses predict how many units they'll sell at different price levels. In this exercise, the demand function is \(p = f(x) = -0.1 x^{2} - x + 40\). This indicator allows seeing how the price (\(p\)) adjusts as the demand in units (\(x\)) changes.

The negative coefficient of the square term \(-0.1 x^{2}\) suggests a nonlinear relationship where increasing demand decreases price, indicating inverse demand behavior typical in market settings. This function is quadratic, depicting more complex price changes than simple linear functions.

Mathematically understanding this helps in decision-making processes like setting optimal price points to maximize revenue or market penetration.
Power Rule
The power rule is a fundamental technique in calculus, especially useful when dealing with polynomials. It simplifies the process of differentiation, making it easy to find derivatives of functions. The rule states that if you have a term \(x^n\), its derivative is \(nx^{n-1}\).

For example, in the given demand function, applying the power rule to \(-0.1 x^2\):
  • Derivative of \(-0.1 x^2\) is \(-0.2 x^1\) or \(-0.2x\).
Similarly, the derivative of \(-x\) (\(x^1\)) becomes \(-1\), and the constant \(40\) has no \(x\) to differentiate, so it drops out to 0.

Using the power rule quickly handles polynomial differentiation needed to find how prices in the tent demand function change with varying sales.

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Most popular questions from this chapter

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