Answer:
1. 375 adult tickets and 125 children's tickets
2. 10x +40y ≤ 20,000; 15x +10y ≤ 17,000
Explanation:
1.
For problems involving mixtures or combinations of two kinds of objects, it is often convenient to choose a variable to represent the object contributing the most to the mix. Here, the adult tickets contribute more to revenue than children's tickets, so we'll use "a" to represent the number of adult tickets sold. Then the number of children's tickets sold is (500-a), since the total is 500 tickets.
The revenue from ticket sales is said to be ...
7.50a + 4.00(500-a) = 3312.50
3.50a +2000 = 3312.50 . . . . eliminate parentheses, collect terms
3.50a = 1312.50 . . . . . . . . . . . subtract 2000
1312.50/3.50 = a = 375 . . . . . divide by the coefficient of "a"
Then the number of children's tickets sold is
(500-a) = 500-375 = 125
375 adult tickets and 125 children's tickets were sold.
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2.
You are given two relationships that must be met. One inequality can be written for each one.
Up to 20,000 ft² of space is available
10x + 40y ≤ 20,000 . . . . . total the area taken by each product
Storage cost per day cannot exceed $17,000
15x + 10y ≤ 17,000 . . . . . . total the storage costs for each product