a bookstore is placing an order for new books and magazines each book costs 15$ and each magazine cost 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 feet of space. the store has only 50 square feet of display space avaiable for the new items the following system of inequalities can be used to determine possible combinations of books x and magazines y the store can order

Question

15x+5y ≤ 1000
2x+y ≤ 50
what combination of books and magazines can the bookstore order

Answer 1

To find the possible combinations of books (x) and magazines (y) that the bookstore can order based on the given inequalities, we need to solve the system of inequalities:

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

We can start by rewriting each inequality in a more helpful form:

1. Cost Constraint

\[ 15x + 5y \leq 1000 \] Dividing the entire inequality by 5: \[ 3x + y \leq 200 \] This can be rewritten as: \[ y \leq 200 - 3x \]

2. Space Constraint

\[ 2x + y \leq 50 \] This can be rewritten as: \[ y \leq 50 - 2x \]

Finding Intersection Points

Now we have two equations to find their intersection points:

1. \( y = 200 - 3x \)
2. \( y = 50 - 2x \)

Setting them equal to find the intersection: \[ 200 - 3x = 50 - 2x \]

Solving for \( x \): \[ 200 - 50 = 3x - 2x \] \[ 150 = x \]

Substituting \( x = 150 \) into one of the equations to find \( y \): Using \( y = 50 - 2x \): \[ y = 50 - 2(150) \] \[ y = 50 - 300 \] \[ y = -250 \] (which is not feasible, as we cannot have negative magazines).

Hence, these lines don't intersect within a feasible region. Now we need to check the boundaries:

Boundary Values of x and y

1. From the cost constraint \( 15x + 5y = 1000 \):

   • If \( x = 0 \): \[ 5y = 1000 \Rightarrow y = 200 \]
   • If \( y = 0 \): \[ 15x = 1000 \Rightarrow x = \frac{1000}{15} \approx 66.67 \] (not a whole number)

2. From the space constraint \( 2x + y = 50 \):

   • If \( x = 0 \): \[ y = 50 \]
   • If \( y = 0 \): \[ 2x = 50 \Rightarrow x = 25 \]

Finding Feasible Region Vertices

Now we can use the maximum feasible points calculated from the inequalities.

1. \( (0, 50) \) - Only magazines
2. \( (25, 0) \) - Only books

Check Capabilities with Input Values

With the values calculated and the constraints verified:

• At \( (0, 50) \): \( 15(0) + 5(50) = 250 \) (meets the cost)
• At \( (25, 0) \): \( 15(25) + 5(0) = 375 \) (meets the cost)

Conclusion

The bookstore can order:

• Up to 50 magazines (0 books),
• Up to 25 books (0 magazines).

However, combinations can be derived (like \( (20, 10) \), \( (10, 25) \) etc.) by solving the constraints but must always adhere to \( 15x + 5y \leq 1000 \) and \( 2x + y \leq 50 \).

Possible Combination

So, feasible combinations can be any values (whole numbers) in this feasible range satisfying both inequalities:

For example:

• If \( x = 10 \): \( 15(10) + 5y \leq 1000 \Rightarrow y \leq 100 \), and \( 2(10) + y \leq 50 \Rightarrow y \leq 30 \). So, maximum \( y = 30 \).
• You can evaluate other combinations like that within your constraints!

To summarize, keep in mind that the solutions must lie within grid lines defined by those equations' intersections.