12. The revenue of a company R as a function of the number of items produced r is given by the function R(x) = -0.5x² + 120 - 500 How many items should the company produce to maximize their revenue? What is that maximum revenue?
12. The revenue of a company R as a function of the number of items produced r is given by the function R(x) = -0.5x² + 120 - 500 How many items should the company produce to maximize their revenue? What is that maximum revenue?
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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![**Problem Statement:**
12. The revenue of a company \( R \) as a function of the number of items produced \( x \) is given by the function:
\[ R(x) = -0.5x^2 + 120x - 500 \]
**Question:**
How many items should the company produce to maximize their revenue? What is that maximum revenue?
**Explanation:**
The function \( R(x) = -0.5x^2 + 120x - 500 \) is a quadratic equation in the standard form \( ax^2 + bx + c \), where \( a = -0.5 \), \( b = 120 \), and \( c = -500 \).
To find the maximum revenue, identify the vertex of the parabola represented by the quadratic function. The x-coordinate of the vertex of a parabola is given by:
\[ x = -\frac{b}{2a} \]
Substituting the values:
\[ x = -\frac{120}{2 \times -0.5} = -\frac{120}{-1} = 120 \]
The company should produce 120 items to maximize revenue.
To find the maximum revenue, substitute \( x = 120 \) back into the revenue function:
\[ R(120) = -0.5(120)^2 + 120(120) - 500 \]
\[ R(120) = -0.5(14400) + 14400 - 500 \]
\[ R(120) = -7200 + 14400 - 500 \]
\[ R(120) = 7200 - 500 \]
\[ R(120) = 6700 \]
Thus, the maximum revenue is 6700.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffa413bcf-f4d4-4563-8c5c-bfbbe2a7d62c%2F70323aac-17f6-4c08-882e-27aeae4600e0%2F6q861ri_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
12. The revenue of a company \( R \) as a function of the number of items produced \( x \) is given by the function:
\[ R(x) = -0.5x^2 + 120x - 500 \]
**Question:**
How many items should the company produce to maximize their revenue? What is that maximum revenue?
**Explanation:**
The function \( R(x) = -0.5x^2 + 120x - 500 \) is a quadratic equation in the standard form \( ax^2 + bx + c \), where \( a = -0.5 \), \( b = 120 \), and \( c = -500 \).
To find the maximum revenue, identify the vertex of the parabola represented by the quadratic function. The x-coordinate of the vertex of a parabola is given by:
\[ x = -\frac{b}{2a} \]
Substituting the values:
\[ x = -\frac{120}{2 \times -0.5} = -\frac{120}{-1} = 120 \]
The company should produce 120 items to maximize revenue.
To find the maximum revenue, substitute \( x = 120 \) back into the revenue function:
\[ R(120) = -0.5(120)^2 + 120(120) - 500 \]
\[ R(120) = -0.5(14400) + 14400 - 500 \]
\[ R(120) = -7200 + 14400 - 500 \]
\[ R(120) = 7200 - 500 \]
\[ R(120) = 6700 \]
Thus, the maximum revenue is 6700.
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