Answer: The feasible region for the number of tickets that can be given away is defined by the inequalities \( x + y \geq 20 \) and \( 10x + 20y \leq 300 \).
Explanation: This problem involves linear inequalities and graphing a feasible region. The goal is to determine how many tickets of each kind (costing $10 and $20) can be given away while satisfying the conditions of giving away at least 20 tickets and not exceeding a total cost of $300.
Steps:
- Define Variables:
- Let \( x \) be the number of $10 tickets.
- Let \( y \) be the number of $20 tickets.
- Set Up Inequalities:
- The total number of tickets must be at least 20:
\[
x + y \geq 20
\]
- The total cost of the tickets must not exceed $300:
\[
10x + 20y \leq 300
\]
- Simplify the Cost Inequality:
- Divide the entire inequality by 10 to simplify:
\[
x + 2y \leq 30
\]
- Graph the Inequalities:
- Plot the line \( x + y = 20 \). This line represents the boundary where exactly 20 tickets are given away.
- Plot the line \( x + 2y = 30 \). This line represents the boundary where the total cost is exactly $300.
- Determine the Feasible Region:
- The feasible region is the area on the graph where both conditions are satisfied. It is the intersection of the regions defined by the inequalities \( x + y \geq 20 \) and \( x + 2y \leq 30 \).
- Find Intersections:
- Solve the system of equations to find intersection points:
- \( x + y = 20 \)
- \( x + 2y = 30 \)
- Subtract the first equation from the second:
\[
(x + 2y) - (x + y) = 30 - 20
\]
\[
y = 10
\]
- Substitute \( y = 10 \) into \( x + y = 20 \):
\[
x + 10 = 20 \Rightarrow x = 10
\]
- Intersection point: \( (10, 10) \).
- Check Boundary Points:
- \( x = 0 \): \( y = 20 \) (from \( x + y = 20 \))
- \( y = 0 \): \( x = 30 \) (from \( x + 2y = 30 \))
- Conclusion:
- The feasible region is bounded by the points \( (0, 20) \), \( (10, 10) \), and \( (30, 0) \).
- The solution set includes all integer points within and on the boundary of this region, representing combinations of tickets that satisfy both constraints.