To determine which combinations of books (x) and magazines (y) the bookstore can order, we need to evaluate each option against the given system of inequalities:
-
Budget Constraint:
\( 15x + 5y \leq 1000 \) -
Space Constraint:
\( 2x + y \leq 50 \)
Let's evaluate each combination:
Option 1: 20 books and 15 magazines (x = 20, y = 15)
-
For the budget:
\( 15(20) + 5(15) = 300 + 75 = 375 \)
\( 375 \leq 1000 \) (satisfied) -
For the space:
\( 2(20) + 15 = 40 + 15 = 55 \)
\( 55 \leq 50 \) (not satisfied)
Option 2: 50 books and 0 magazines (x = 50, y = 0)
-
For the budget:
\( 15(50) + 5(0) = 750 + 0 = 750 \)
\( 750 \leq 1000 \) (satisfied) -
For the space:
\( 2(50) + 0 = 100 + 0 = 100 \)
\( 100 \leq 50 \) (not satisfied)
Option 3: 10 books and 30 magazines (x = 10, y = 30)
-
For the budget:
\( 15(10) + 5(30) = 150 + 150 = 300 \)
\( 300 \leq 1000 \) (satisfied) -
For the space:
\( 2(10) + 30 = 20 + 30 = 50 \)
\( 50 \leq 50 \) (satisfied)
Option 4: 40 books and 5 magazines (x = 40, y = 5)
-
For the budget:
\( 15(40) + 5(5) = 600 + 25 = 625 \)
\( 625 \leq 1000 \) (satisfied) -
For the space:
\( 2(40) + 5 = 80 + 5 = 85 \)
\( 85 \leq 50 \) (not satisfied)
Summary of Results
- Option 1: Not valid (space)
- Option 2: Not valid (space)
- Option 3: Valid (both constraints satisfied)
- Option 4: Not valid (space)
The only valid combination is:
10 books and 30 magazines.