Answer:
Two possible combinations are;
1) 200 single and 30 couples
2) 150 single and 57 couples
Explanation:
The given parameter in the question are;
The cost of ticket per person = $5
The cost of tickets per couple = $8
The minimum worth of tickets to be sold = $1,200
The maximum number of students that can fit in the gym = 500
Let 'x' represent the number of single persons that attend the dance and let 'y' represent the number of couple that attend, we have;
5·x + 8·y ≥ 1,200
x + y ≤ 500
Making 'y' the subject of both equations gives;
y ≥ 150 - 5·x/8
y ≤ 500 - x
From the graph of the inequalities created with Microsoft Excel, we have;
When x = 200, and y = 30, we get;
5 × 200 + 8 × 30 = 1240 > 1200
200 + 30 × 2 = 460 < 500
We also have;
x = 150, and y = 57, we get
Equating both values of 'y', we get;
5 × 150 + 8 × 57 = 1206 > 1200
150 + 57 × 2 = 264 < 500
Therefore, from the acceptable region in the graph, we have the following two possible combinations
1) 200 single and 30 couples
2) 150 single and 57 couples