Santiago is working at his mom’s day care this summer. His mom has asked him to check out the

Question

library book sale to look for books they might be able to use in the day care. When he gets to the
book sale, he sees that board books are $1.50 and soft cover books are $0.50. He picks out a variety of books and spends $20.

What are the constraints of the situation? Describe them.

Answer 1

In this scenario, Santiago is trying to maximize his selection of books while adhering to certain constraints. The constraints can be defined based on the following factors:

1. Budget Constraint: Santiago has a total budget of $20. This means the total amount he spends must not exceed $20. If we let \( x \) represent the number of board books and \( y \) represent the number of soft cover books, the budget constraint can be expressed mathematically as: \[ 1.50x + 0.50y \leq 20 \]

2. Non-negativity Constraint: Santiago cannot buy a negative number of books. Therefore, the number of board books and soft cover books must be non-negative. This can be represented as: \[ x \geq 0 \quad \text{and} \quad y \geq 0 \]

3. Whole Number Constraint: Since books must be purchased in whole numbers, \( x \) and \( y \) must be integers. This implies that: \[ x \in \mathbb{Z}^+ \quad \text{and} \quad y \in \mathbb{Z}^+ \]

4. Types of Books Available: While not a numerical constraint, it's also understood that he is limited to purchasing only board and soft cover books, meaning he can't buy hardcover books or other forms of media.

These constraints together define the parameters within which Santiago must operate while making his selections at the book sale.

Answer 2

Santiago is working at his mom’s day care this summer. His mom has asked him to check out the
library book sale to look for books they might be able to use in the day care. When he gets to the
book sale, he sees that board books are $1.50 and soft cover books are $0.50. He picks out a variety of books and spends $20.

Use your graph to help you make a list of all viable solutions to the problem.

Answer 3

To find all viable solutions to Santiago's problem, we first need to set up the inequalities representing the constraints, as stated previously.

1. Budget Constraint: \[ 1.50x + 0.50y \leq 20 \]

2. Non-negativity Constraint: \[ x \geq 0 \quad \text{and} \quad y \geq 0 \]

3. Whole Number Constraint: \( x \) and \( y \) should be integers.

Now let's rewrite the budget constraint for easier handling:

\[ 1.50x + 0.50y \leq 20 \implies 3x + y \leq 40 \]

Now, we can solve for \( y \) in terms of \( x \):

\[ y \leq 40 - 3x \]

Next, we will determine the integer solutions to this inequality. Since \( x \) and \( y \) must be non-negative integers, let's consider the possible values of \( x \):

Calculate Possible Values of \( y \) for Each Integer \( x \):

1. If \( x = 0 \): \[ y \leq 40 - 3(0) = 40 \implies y = 0, 1, 2, \ldots, 40 \quad (41 \text{ solutions}) \]

2. If \( x = 1 \): \[ y \leq 40 - 3(1) = 37 \implies y = 0, 1, 2, \ldots, 37 \quad (38 \text{ solutions}) \]

3. If \( x = 2 \): \[ y \leq 40 - 3(2) = 34 \implies y = 0, 1, 2, \ldots, 34 \quad (35 \text{ solutions}) \]

4. If \( x = 3 \): \[ y \leq 40 - 3(3) = 31 \implies y = 0, 1, 2, \ldots, 31 \quad (32 \text{ solutions}) \]

5. If \( x = 4 \): \[ y \leq 40 - 3(4) = 28 \implies y = 0, 1, 2, \ldots, 28 \quad (29 \text{ solutions}) \]

6. If \( x = 5 \): \[ y \leq 40 - 3(5) = 25 \implies y = 0, 1, 2, \ldots, 25 \quad (26 \text{ solutions}) \]

7. If \( x = 6 \): \[ y \leq 40 - 3(6) = 22 \implies y = 0, 1, 2, \ldots, 22 \quad (23 \text{ solutions}) \]

8. If \( x = 7 \): \[ y \leq 40 - 3(7) = 19 \implies y = 0, 1, 2, \ldots, 19 \quad (20 \text{ solutions}) \]

9. If \( x = 8 \): \[ y \leq 40 - 3(8) = 16 \implies y = 0, 1, 2, \ldots, 16 \quad (17 \text{ solutions}) \]

10. If \( x = 9 \): \[ y \leq 40 - 3(9) = 13 \implies y = 0, 1, 2, \ldots, 13 \quad (14 \text{ solutions}) \]

11. If \( x = 10 \): \[ y \leq 40 - 3(10) = 10 \implies y = 0, 1, 2, \ldots, 10 \quad (11 \text{ solutions}) \]

12. If \( x = 11 \): \[ y \leq 40 - 3(11) = 7 \implies y = 0, 1, 2, \ldots, 7 \quad (8 \text{ solutions}) \]

13. If \( x = 12 \): \[ y \leq 40 - 3(12) = 4 \implies y = 0, 1, 2, \ldots, 4 \quad (5 \text{ solutions}) \]

14. If \( x = 13 \): \[ y \leq 40 - 3(13) = 1 \implies y = 0, 1 \quad (2 \text{ solutions}) \]

15. If \( x = 14 \): \[ y \leq 40 - 3(14) = -2 \implies y \text{ cannot be non-negative} \quad (0 \text{ solutions}) \]

Summary of Viable Solutions:

Now, we can summarize the viable solutions as pairs \( (x, y) \):

• For \( x = 0 \): \( (0, 0), (0, 1), (0, 2), ..., (0, 40) \)
• For \( x = 1 \): \( (1, 0), (1, 1), (1, 2), ..., (1, 37) \)
• For \( x = 2 \): \( (2, 0), (2, 1), (2, 2), ..., (2, 34) \)
• For \( x = 3 \): \( (3, 0), (3, 1), (3, 2), ..., (3, 31) \)
• For \( x = 4 \): \( (4, 0), (4, 1), (4, 2), ..., (4, 28) \)
• For \( x = 5 \): \( (5, 0), (5, 1), ..., (5, 25) \)
• For \( x = 6 \): \( (6, 0), (6, 1), ..., (6, 22) \)
• For \( x = 7 \): \( (7, 0), (7, 1), ..., (7, 19) \)
• For \( x = 8 \): \( (8, 0), (8, 1), ..., (8, 16) \)
• For \( x = 9 \): \( (9, 0), (9, 1), ..., (9, 13) \)
• For \( x = 10 \): \( (10, 0), (10, 1), ..., (10, 10) \)
• For \( x = 11 \): \( (11, 0), (11, 1), ..., (11, 7) \)
• For \( x = 12 \): \( (12, 0), (12, 1), ..., (12, 4) \)
• For \( x = 13 \): \( (13, 0), (13, 1) \)

Each of these pairs represents a feasible solution for how many board books (\( x \)) and how many soft cover books (\( y \)) Santiago can purchase without exceeding his budget of $20.

Answer 4

Santiago is working at his mom’s day care this summer. His mom has asked him to check out the
library book sale to look for books they might be able to use in the day care. When he gets to the
book sale, he sees that board books are $1.50 and soft cover books are $0.50. He picks out a variety of books and spends $20.

Suppose Santiago bought 34 books. Can you figure out how many of each type of book he
bought? Explain.

Answer 5

To determine how many board books and softcover books Santiago bought, we need to set up a system of equations based on the information provided:

1. Total Number of Books Equation: If \( x \) is the number of board books and \( y \) is the number of softcover books, we have: \[ x + y = 34 \]

2. Total Cost Equation: The cost of board books is $1.50 and the cost of softcover books is $0.50, and the total amount spent is $20: \[ 1.50x + 0.50y = 20 \]

Now, we can rearrange the first equation to express \( y \) in terms of \( x \): \[ y = 34 - x \]

Next, we'll substitute this expression for \( y \) into the second equation:

\[ 1.50x + 0.50(34 - x) = 20 \]

Now, simplify this equation:

\[ 1.50x + 17 - 0.50x = 20 \]

Combine like terms:

\[ (1.50 - 0.50)x + 17 = 20 \] \[ 1.00x + 17 = 20 \]

Next, solve for \( x \):

\[ 1.00x = 20 - 17 \] \[ 1.00x = 3 \] \[ x = 3 \]

Now, substitute \( x = 3 \) back into the equation for \( y \):

\[ y = 34 - x = 34 - 3 = 31 \]

Conclusion:

Santiago bought:

• 3 board books
• 31 softcover books

We can verify this by checking the total cost:

• Total cost for board books: \( 3 \times 1.50 = 4.50 \)
• Total cost for softcover books: \( 31 \times 0.50 = 15.50 \)
• Total cost: \( 4.50 + 15.50 = 20.00 \)

The calculations confirm that the number of books and total cost are consistent with the problem's constraints.