Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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![**Understanding the Relationship Between Advertising Expenditure and Sales**
The relationship between sales \(S\) and the amount \(x\) and \(y\) spent on two advertising media is described by the following equation:
\[ S = \frac{200x}{5 + y} + \frac{100y}{10 + y} \]
### Explanation of Terms:
- \(S\): The total sales generated.
- \(x\): The amount spent on the first advertising medium.
- \(y\): The amount spent on the second advertising medium.
### Given Constraints:
- Net profit is defined as \(\frac{1}{5}\) of sales minus the costs of advertising.
- The total advertising cost must not exceed the budget set at 25.
### Objective:
Determine how the advertising budget of 25 should be optimally allocated between the two media (\(x\) and \(y\)) to maximize net profit.
### Steps to Maximize Net Profit:
1. **Formulate the Net Profit Equation:**
The net profit \(P\) can be expressed as:
\[ P = \frac{1}{5}S - (x + y) \]
2. **Substitute the Sales Equation:**
Replace \(S\) in the net profit equation using the given sales relationship:
\[ S = \frac{200x}{5 + y} + \frac{100y}{10 + y} \]
Hence,
\[ P = \frac{1}{5} \left( \frac{200x}{5 + y} + \frac{100y}{10 + y} \right) - (x + y) \]
3. **Implement the Budget Constraint:**
According to the constraint:
\[ x + y \leq 25 \]
By simultaneously optimizing \(P\) while ensuring that the advertising cost does not exceed the budget \(x + y \leq 25\), you can determine the optimal allocation for \(x\) and \(y\) that maximizes the net profit.
This problem typically requires use of calculus or numerical methods for optimization, involving techniques in differential calculus or linear programming to find the values of \(x\) and \(y\) that provide the highest net profit within the given constraints.](https://content.bartleby.com/qna-images/question/1beaffe7-9d7a-42b9-9107-dccba3680f89/303e70a8-69fb-4336-a0ca-e58fa8b8b37c/spn87fh_processed.jpeg)
Transcribed Image Text:**Understanding the Relationship Between Advertising Expenditure and Sales**
The relationship between sales \(S\) and the amount \(x\) and \(y\) spent on two advertising media is described by the following equation:
\[ S = \frac{200x}{5 + y} + \frac{100y}{10 + y} \]
### Explanation of Terms:
- \(S\): The total sales generated.
- \(x\): The amount spent on the first advertising medium.
- \(y\): The amount spent on the second advertising medium.
### Given Constraints:
- Net profit is defined as \(\frac{1}{5}\) of sales minus the costs of advertising.
- The total advertising cost must not exceed the budget set at 25.
### Objective:
Determine how the advertising budget of 25 should be optimally allocated between the two media (\(x\) and \(y\)) to maximize net profit.
### Steps to Maximize Net Profit:
1. **Formulate the Net Profit Equation:**
The net profit \(P\) can be expressed as:
\[ P = \frac{1}{5}S - (x + y) \]
2. **Substitute the Sales Equation:**
Replace \(S\) in the net profit equation using the given sales relationship:
\[ S = \frac{200x}{5 + y} + \frac{100y}{10 + y} \]
Hence,
\[ P = \frac{1}{5} \left( \frac{200x}{5 + y} + \frac{100y}{10 + y} \right) - (x + y) \]
3. **Implement the Budget Constraint:**
According to the constraint:
\[ x + y \leq 25 \]
By simultaneously optimizing \(P\) while ensuring that the advertising cost does not exceed the budget \(x + y \leq 25\), you can determine the optimal allocation for \(x\) and \(y\) that maximizes the net profit.
This problem typically requires use of calculus or numerical methods for optimization, involving techniques in differential calculus or linear programming to find the values of \(x\) and \(y\) that provide the highest net profit within the given constraints.
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