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Chapter 8: Problem 24

a. In economics, revenue \(R\) is defined as the amount of money derived from the sale of a product and is equal to the number \(x\) of units sold times the selling price \(p\) of each unit. What is the equation for revenue? b. If the selling price is given by the equation \(p=-\frac{1}{10} x+20,\) express revenue \(R\) as a function of the number \(x\) of units sold. c. Using technology, plot the function and estimate the number of units that need to be sold to achieve maximum revenue. Then estimate the maximum revenue.

Short Answer

Expert verified
The revenue function is \( R(x) = -\frac{1}{10}x^2 + 20x \). The maximum revenue of 1000 occurs when 100 units are sold.

Step by step solution

01

Define the revenue equation

Revenue, denoted as R, is calculated by multiplying the number of units sold, x, by the price per unit, p. This can be represented by the equation: \[ R = x \times p \]
02

Substitute the given price equation

Given that the price per unit is defined by the equation \( p = -\frac{1}{10}x + 20 \), substitute this into the revenue equation: \[ R = x \times \bigg(-\frac{1}{10}x + 20\bigg) \]
03

Simplify the revenue equation

Expand the equation to express revenue as a function of the number of units sold: \[ R = x \times -\frac{1}{10}x + x \times 20 \] Simplify further: \[ R = -\frac{1}{10} x^2 + 20x \]
04

Rewrite the revenue function

The simplified revenue function is: \[ R(x) = -\frac{1}{10} x^2 + 20x \]
05

Plot the function and find the maximum

Using a graphing tool, plot the function \( R(x) = -\frac{1}{10} x^2 + 20x \). The graph is a downward-opening parabola. The vertex of this parabola represents the maximum revenue. For the quadratic function \( ax^2 + bx + c \), the vertex occurs at \( x = -\frac{b}{2a} \). In this case, \( a = -\frac{1}{10} \) and \( b = 20 \), so: \[ x = -\frac{20}{2 \times -\frac{1}{10}} \] Simplify: \[ x = 100 \] This means the maximum revenue occurs when 100 units are sold.
06

Calculate the maximum revenue

Substitute \( x = 100 \) back into the revenue function to find the maximum revenue: \[ R(100) = -\frac{1}{10}(100)^2 + 20(100) \] Simplify: \[ R(100) = -1000 + 2000 \] \[ R(100) = 1000 \]

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Key Concepts

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

Revenue calculation
Revenue calculation is a fundamental concept in economics. Revenue, denoted as \( R \), is derived from the total amount of money made through the sale of goods or services. This calculation is crucial for businesses to understand their earnings. The basic formula to calculate revenue is quite straightforward. It is given by multiplying the number of units sold \( x \) by the selling price per unit \( p \). Therefore, the equation for revenue can be expressed as:

\[ R = x \times p \]

For example, if you sell 50 units of a product priced at \(10 each, your revenue would be 50 \times 10 = \)500.

This simple relationship helps businesses set sales targets and pricing strategies. By examining how changes in price and sales volume affect revenue, businesses can make informed decisions.
Quadratic functions
Quadratic functions are widely used in economics to model various relationships. A quadratic function is a type of polynomial function that can be expressed in the standard form:

\[ f(x) = ax^2 + bx + c \]

In our exercise, revenue \( R(x) \) is expressed as a quadratic function: \[ R(x) = -\frac{1}{10} x^2 + 20x \]

Quadratic functions graph as parabolas which can open upwards or downwards. In this case, the negative coefficient of \( x^2 \) indicates that the parabola opens downwards. Hence, the function has a maximum value, which is vital for determining the maximum revenue.

Understanding quadratic functions helps in analyzing how different factors influence overall outcomes. For example, in revenue calculations, it shows how revenue grows or declines with sales volume.
Maximum revenue
Identifying the maximum revenue is essential for businesses to maximize profits. In the context of a quadratic revenue function, the maximum revenue occurs at the vertex of the parabola.

The vertex formula for a quadratic function \( ax^2 + bx + c \) is given by:

\[ x = -\frac{b}{2a} \]

Using our revenue function \( R(x) = -\frac{1}{10} x^2 + 20x \), we can find the number of units \( x \) that maximizes revenue:

\[ a = -\frac{1}{10}, b = 20 \]

Plug these values into the vertex formula:

\[ x = -\frac{20}{2 \times -\frac{1}{10}} = 100 \]

This calculation shows that selling 100 units will maximize revenue. To find the maximum revenue, substitute \( x = 100 \) back into the revenue function:

\[ R(100) = -\frac{1}{10}(100)^2 + 20(100) \]

Simplify to get:

\[ R(100) = -1000 + 2000 = 1000 \]

Therefore, the maximum revenue is $1000 when 100 units are sold. This illustrates the importance of quadratic functions in pinpointing peak performance for sales and revenue.

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Most popular questions from this chapter

A Norman window has the shape of a rectangle surmounted by a semicircle of diameter equal to the width of the rectangle (see the accompanying figure). If the perimeter of the window is 20 feet (including the semicircle), what dimensions will admit the most light (maximize the area)? (Hint: Express \(L\) in terms of \(r\). Recall that the circumference of a circle \(=2 \pi r,\) and the area of a circle \(=\pi r^{2},\) where \(r\) is the radius of the circle.)

For the following quadratic functions in vertex form, \(f(x)=a(x-h)^{2}+k,\) determine the values for \(a, h,\) and \(k\) Then compare each to \(f(x)=x^{2},\) and identify which constants represent a stretch/compression factor, or a shift in a particular direction. a. \(p(x)=5(x-4)^{2}-2\) b. \(g(x)=\frac{1}{3}(x+5)^{2}+4\) c. \(h(x)=-0.25\left(x-\frac{1}{2}\right)^{2}+6\) d. \(k(x)=-3(x+4)^{2}-3\)

Let \(f(x)=x^{3}\). a. Write the equation for the new function \(g(x)\) that results from each of the following transformations of \(f(x)\). Explain in words the effect of the transformations. i. \(f(-x)\) iii. \(f(x+2)\) ii. \(-2 f(x)-1\) iv. \(-f(-x)\) b. Sketch by hand the graph of \(f(x)\) and each function in part (a).

If we know the radius and depth of a parabolic reflector, we also know where the focus is. a. Find a generic formula for the focal length \(f\) of a parabolic reflector expressed in terms of its radius \(R\) and depth \(D\). The focal length \(\left|\frac{1}{4 a}\right|\) is the distance between the vertex and the focal point. Assume \(a>0\). b. Under what conditions does \(f=D ?\)

A shot-put athlete releases the shot at a speed of 14 meters per second, at an angle of 45 degrees to the horizontal (ground level). The height \(y\) (in meters above the ground) of the shot is given by the function $$ y=2+x-\frac{1}{20} x^{2} $$ where \(x\) is the horizontal distance the shot has traveled (in meters). a. What was the height of the shot at the moment of release? b. How high is the shot after it has traveled 4 meters horizontally from the release point? 16 meters? c. Find the highest point reached by the shot in its flight. d. Draw a sketch of the height of the shot and indicate how far the shot is from the athlete when it lands.
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