The monthly profit from the sale of a product is given by P = 16x – 0.1x2 – 200 dollars. (a) What level of production maximizes profit? units (b) What is the maximum possible profit?
The monthly profit from the sale of a product is given by P = 16x – 0.1x2 – 200 dollars. (a) What level of production maximizes profit? units (b) What is the maximum possible profit?
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![The monthly profit from the sale of a product is given by the equation:
\[ P = 16x - 0.1x^2 - 200 \] dollars.
(a) What level of production maximizes profit?
\[ \_\_\_\_ \] units
(b) What is the maximum possible profit?
\[ \$\_\_\_\_ \]
**Explanation:**
- The equation represents a quadratic function where \( P \) is the profit, and \( x \) is the number of units produced.
- To find the level of production that maximizes profit, determine the vertex of the parabola, since this is a downward-opening parabola (the coefficient of \( x^2 \) is negative).
- The maximum profit can be calculated by substituting the value of \( x \) that maximizes profit back into the profit equation.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc2ed13a8-94cc-42b1-92cb-21374826aa79%2Fb57ea8e3-8233-41c1-9b85-90a4cd6678a7%2Fyad0pu_processed.png&w=3840&q=75)
Transcribed Image Text:The monthly profit from the sale of a product is given by the equation:
\[ P = 16x - 0.1x^2 - 200 \] dollars.
(a) What level of production maximizes profit?
\[ \_\_\_\_ \] units
(b) What is the maximum possible profit?
\[ \$\_\_\_\_ \]
**Explanation:**
- The equation represents a quadratic function where \( P \) is the profit, and \( x \) is the number of units produced.
- To find the level of production that maximizes profit, determine the vertex of the parabola, since this is a downward-opening parabola (the coefficient of \( x^2 \) is negative).
- The maximum profit can be calculated by substituting the value of \( x \) that maximizes profit back into the profit equation.
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