Question

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 possible combinations of books, x, and magazines, y, the store can order:
15x + 5y ≤ 1000
2x + y ≤ 50
Which combinations of books, x, and magazines, y, can the bookstore order?

Answer 1

To determine the possible combinations of books \(x\) and magazines \(y\) that the bookstore can order, we need to analyze the given system of inequalities:

1. The budget constraint:
\[
15x + 5y \leq 1000
\]
2. The space constraint:
\[
2x + y \leq 50
\]

Additionally, both \(x\) and \(y\) must be non-negative:
\[
x \geq 0 \quad \text{and} \quad y \geq 0
\]

### Step 1: Graph the inequalities

We'll rewrite each inequality to express \(y\) in terms of \(x\) to make graphing easier.

**For the budget inequality:**
\[
15x + 5y \leq 1000
\]
Subtract \(15x\) from both sides:
\[
5y \leq 1000 - 15x
\]
Divide by 5:
\[
y \leq 200 - 3x
\]

**For the space inequality:**
\[
2x + y \leq 50
\]
Subtract \(2x\) from both sides:
\[
y \leq 50 - 2x
\]

### Step 2: Identify the vertices of the feasible region

Next, we will find the points of intersection of the lines \(y = 200 - 3x\) and \(y = 50 - 2x\):
Set the two equations equal to each other:
\[
200 - 3x = 50 - 2x
\]
Rearranging gives:
\[
200 - 50 = 3x - 2x
\]
\[
150 = x
\]
Now substitute \(x = 150\) back into one of the equations to find \(y\):
\[
y = 200 - 3(150) = 200 - 450 = -250
\]

Since \(y\) cannot be negative, this intersection point is not within our feasible region.

### Step 3: Find the x-intercepts and y-intercepts

**For \(15x + 5y = 1000\) (budget line):**
- When \(x = 0\):
\[
15(0) + 5y = 1000 \implies y = 200
\]
- When \(y = 0\):
\[
15x + 5(0) = 1000 \implies x = \frac{1000}{15} \approx 66.67
\]

**For \(2x + y = 50\) (space line):**
- When \(x = 0\):
\[
2(0) + y = 50 \implies y = 50
\]
- When \(y = 0\):
\[
2x + 0 = 50 \implies x = 25
\]

### Step 4: Identify the feasible region

We have marked the intercepts of the lines:
- \(y \leq 200 - 3x\) crosses the axes at (0, 200) and (66.67, 0).
- \(y \leq 50 - 2x\) crosses the axes at (0, 50) and (25, 0).

### Step 5: Check the corner points

The feasible region is bound by the intercepts:
1. (0, 0)
2. (0, 50)
3. (25, 0)
4. (25, 50) - Check whether this point lies in both inequalities

Substituting (25, 0):
- For budget: \(15(25) + 5(0) = 375 \leq 1000\) (True)
- For space: \(2(25) + 0 = 50 \leq 50\) (True)

### Conclusion

Thus, the bookstore can order combinations of books and magazines satisfying:
- \(0 \leq x \leq 25\)
- \(0 \leq y \leq 50\)
- Checking additional integer combinations from these limits can give specific combinations for the order.

For example, some combinations could be:
- 0 books and 50 magazines
- 1 book and 48 magazines
- 25 books and 0 magazines
- 10 books and 30 magazines, etc.

In general, \( (x, y) \) must satisfy both inequalities while both being non-negative, and integers for quantities.