A radio station is giving away tickets to a play. They plan to give away tickets for seats that cost $10 and $20. They want to give away at least 20 tickets. The total cost of all the tickets they give away can be no more than $280. Write the inequalities that represent this situation and graph.
The Correct Answer and Explanation is:
To solve this problem, we need to create a system of inequalities based on the given conditions. Let’s define the variables:
- Let x represent the number of $10 tickets.
- Let y represent the number of $20 tickets.
Step 1: Creating the inequalities
The radio station is giving away at least 20 tickets, which leads to the inequality:
[ x + y \geq 20 ]
This means the total number of tickets given away must be at least 20.
Next, the total cost of the tickets can be no more than $280. The cost of $10 tickets is ( 10x ), and the cost of $20 tickets is ( 20y ). Therefore, the inequality for the cost of tickets is:
[ 10x + 20y \leq 280 ]
Step 2: Simplifying the cost inequality
We can simplify the inequality ( 10x + 20y \leq 280 ) by dividing the entire inequality by 10:
[ x + 2y \leq 28 ]
Step 3: Non-negativity constraints
Additionally, since the number of tickets cannot be negative, we add the non-negativity constraints:
[ x \geq 0 \quad \text{and} \quad y \geq 0 ]
Step 4: Graphing the inequalities
The system of inequalities is:
- ( x + y \geq 20 )
- ( x + 2y \leq 28 )
- ( x \geq 0 )
- ( y \geq 0 )
To graph these inequalities:
- Graph ( x + y = 20 ): This is a straight line with intercepts at ( (20, 0) ) and ( (0, 20) ). The region above this line satisfies ( x + y \geq 20 ).
- Graph ( x + 2y = 28 ): This is a straight line with intercepts at ( (28, 0) ) and ( (0, 14) ). The region below this line satisfies ( x + 2y \leq 28 ).
- The constraints ( x \geq 0 ) and ( y \geq 0 ) limit the solution to the first quadrant.
Step 5: Identifying the solution region
The solution region is the area where all inequalities overlap. This region represents the feasible number of $10 and $20 tickets that can be given away while satisfying both the minimum ticket requirement and the cost limitation.
Conclusion
The solution to this problem is the set of points in the feasible region on the graph. These points represent combinations of $10 and $20 tickets that meet the criteria. To determine the exact number of tickets for specific scenarios, we could test points within this region (like ( x = 10, y = 10 )) to see if they satisfy the inequalities.