A bookstore is placing an order for new books and magazines each book costs $15 and each magazine costs $5 the store can spend at most $1,000 on these books and magazines additionally each book takes up 2 square feet of space and each magazine takes up 1 square foot of space the store has only 50 square feet of display space available for the new items the following system of inequalities can be used to determine the possible combinations of books x and magazines y the store can order.

Question

15x + 5y <1000
2x + y < 50
Which combinations of books, x, and magazines, y, can the bookstore order?
A. 50 books and 0 magazines
B. 20 books and 15 magazines
C. 40 books and 5 magazines
D. 10 books and 30 magazines

Answer 1

To determine the valid combinations of books (x) and magazines (y) that the bookstore can order, we need to check each option against the given inequalities:

1. Cost Inequality: \(15x + 5y \leq 1000\)
2. Space Inequality: \(2x + y \leq 50\)

Let's analyze each option.

Option A: 50 books and 0 magazines (x = 50, y = 0)

• Cost: \(15(50) + 5(0) = 750 \leq 1000\) (satisfied)
• Space: \(2(50) + 0 = 100 \not\leq 50\) (not satisfied)

Option A is NOT valid.

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Option B: 20 books and 15 magazines (x = 20, y = 15)

• Cost: \(15(20) + 5(15) = 300 + 75 = 375 \leq 1000\) (satisfied)
• Space: \(2(20) + 15 = 40 + 15 = 55 \not\leq 50\) (not satisfied)

Option B is NOT valid.

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Option C: 40 books and 5 magazines (x = 40, y = 5)

• Cost: \(15(40) + 5(5) = 600 + 25 = 625 \leq 1000\) (satisfied)
• Space: \(2(40) + 5 = 80 + 5 = 85 \not\leq 50\) (not satisfied)

Option C is NOT valid.

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Option D: 10 books and 30 magazines (x = 10, y = 30)

• Cost: \(15(10) + 5(30) = 150 + 150 = 300 \leq 1000\) (satisfied)
• Space: \(2(10) + 30 = 20 + 30 = 50 \leq 50\) (satisfied)

Option D is valid.

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Conclusion: the only valid combination the bookstore can order is:

D. 10 books and 30 magazines