Donna De Paul is raising money for the homeless. She discovers that each church group requires 2 hours of letter writing and 1 hour of​ follow-up, while for each labor union she needs 2 hours of letter writing and 3 hours of​ follow-up. Donna can raise ​$150 from each church group and ​$175 from each union​ local, and she has a maximum of 16 hours of letter writing and a maximum of 12 hours of​ follow-up available per month. Determine the most profitable mixture of groups she should contact and the most money she can raise in a month.

This problem can be solved using linear programming, as it involves optimizing an objective function (the amount of money Donna can raise) under several constraints (the available hours for letter writing and follow-up).

Let's define the variables:

- Let $ x $ represent the number of church groups Donna contacts.
- Let $ y $ represent the number of labor union locals Donna contacts.

### Objective Function
Donna wants to maximize the total amount of money she can raise. The objective function is based on the amounts she can raise from each group:
\[
\text{Maximize:} \ Z = 150x + 175y
\]
Where:
- $ 150x $ represents the money raised from church groups.
- $ 175y $ represents the money raised from labor union locals.

### Constraints
There are two main constraints, based on her available time for letter writing and follow-up:

1. **Letter Writing Constraint**: Each church group requires 2 hours of letter writing, and each union local also requires 2 hours of letter writing. Donna has a maximum of 16 hours available for letter writing:
\[
2x + 2y \leq 16
\]
Simplifying this gives:
\[
x + y \leq 8
\]

2. **Follow-Up Constraint**: Each church group requires 1 hour of follow-up, and each union local requires 3 hours of follow-up. Donna has a maximum of 12 hours available for follow-up:
\[
x + 3y \leq 12
\]

Additionally, the number of groups must be non-negative:
\[
x \geq 0, \quad y \geq 0
\]

### Solution

We now solve this system of inequalities graphically or algebraically. Let's solve it algebraically.

1. From the constraint $ x + y \leq 8 $, solve for $ x $:
\[
x = 8 - y
\]

2. Substitute $ x = 8 - y $ into the second constraint $ x + 3y \leq 12 $:
\[
8 - y + 3y \leq 12
\]
\[
8 + 2y \leq 12
\]
\[
2y \leq 4 \quad \Rightarrow \quad y \leq 2
\]

Thus, $ y \leq 2 $. Now, substitute $ y = 2 $ into $ x + y \leq 8 $:
\[
x + 2 \leq 8 \quad \Rightarrow \quad x \leq 6
\]

Thus, the solution must lie within the region where $ x \leq 6 $ and $ y \leq 2 $.

### Vertices of the Feasible Region
The feasible region is bounded by the points (0,0), (6,2), (8,0), and (0,4).

- At $ (0, 0) $, $ Z = 150(0) + 175(0) = 0 $
- At $ (6, 2) $, $ Z = 150(6) + 175(2) = 900 + 350 = 1250 $
- At $ (8, 0) $, $ Z = 150(8) + 175(0) = 1200 $
- At $ (0, 4) $, $ Z = 150(0) + 175(4) = 700 $

### Conclusion
The maximum occurs at $ (6, 2) $, meaning Donna should contact 6 church groups and 2 labor union locals to maximize her profit. The maximum amount of money she can raise is \$1250.

---

Would you like more details or have any questions about the solution?

Here are 5 related questions you might find interesting:
1. What would happen if Donna had 20 hours available for letter writing instead of 16?
2. How would the solution change if labor unions required 4 hours of follow-up?
3. What if church groups only raised \$100 each instead of \$150?
4. How can we approach this problem using a graphical method?
5. Can this problem be solved using simplex algorithm techniques?

**Tip**: Linear programming is a powerful tool for optimizing resources and maximizing profit, and it's widely used in various industries like manufacturing, logistics, and finance.

Solve a linear programming problem to determine how Donna De Paul can maximize her fundraising for the homeless by contacting church groups and labor unions. Learn how to optimize her time with constraints on letter writing and follow-up, using algebraic methods and inequalities. Suitable for students in Grades 10-12.

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