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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.
What are the constraints of the situation?

5 answers

To determine the constraints of Santiago's book buying situation, we need to define the variables involved and the limits imposed by the scenario. Let's define two variables:

  • Let \( x \) represent the number of board books he buys.
  • Let \( y \) represent the number of soft cover books he buys.

Given the prices:

  • Board books cost $1.50 each.
  • Soft cover books cost $0.50 each.

Santiago's total spending is $20. This leads to the following equation representing his total expenditure:

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

Additionally, since he cannot purchase a negative number of books, the following constraints apply:

  1. Non-negativity Constraints: \[ x \geq 0 \quad (\text{number of board books}) \] \[ y \geq 0 \quad (\text{number of soft cover books}) \]

So, the complete set of constraints is:

  1. \( 1.50x + 0.50y \leq 20 \) (ensuring he does not exceed his budget)
  2. \( x \geq 0 \) (he cannot buy a negative number of board books)
  3. \( y \geq 0 \) (he cannot buy a negative number of soft cover books)

These constraints define the possible combinations of board books and soft cover books Santiago can purchase within his budget.

can you write it more simple

Sure! Here’s a simpler version of the constraints for Santiago's book buying situation:

  1. Total Money Spent:

    • The total amount he spends on books can be expressed as:
      \[ 1.50x + 0.50y = 20 \]
      where \( x \) is the number of board books and \( y \) is the number of soft cover books.

  2. Non-negativity:

    • He cannot buy negative numbers of books, so:
      \[ x \geq 0 \quad (\text{board books}) \]
      \[ y \geq 0 \quad (\text{soft cover books}) \]

So the constraints are:

  • \( 1.50x + 0.50y = 20 \)
  • \( x \geq 0 \)
  • \( y \geq 0 \)

This means he can buy as many board and soft cover books as long as the total cost is $20 or less and he doesn't buy a negative number of books.

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

To find all viable solutions for Santiago's book-buying scenario using the constraints \(1.50x + 0.50y = 20\), along with the non-negativity conditions \(x \geq 0\) and \(y \geq 0\), we can rearrange the equation to isolate \(y\):

  1. Rearranging the equation: \[ 0.50y = 20 - 1.50x \] \[ y = \frac{20 - 1.50x}{0.50} \] \[ y = 40 - 3x \]

Now let's analyze this equation and determine the viable solutions by plotting it and considering integer values for \(x\) (number of board books) and \(y\) (number of softcover books).

Points on the Line

  1. Finding Intercepts:
    • When \(x = 0\) (Santiago buys only soft cover books):
      \[ y = 40 - 3(0) = 40 \quad \Rightarrow (0, 40) \]
    • When \(y = 0\) (Santiago buys only board books):
      \[ 0 = 40 - 3x \quad \Rightarrow 3x = 40 \quad \Rightarrow x = \frac{40}{3} \approx 13.33 \quad \Rightarrow (13.33, 0) \]

Since x needs to be a whole number, we should check integer values for \(x\) from 0 to 13.

  1. Finding Viable Integer Solutions: Let's list combinations of \(x\) and \(y\) using \(x\) as integers from 0 to 13:

  • If \(x = 0\):
    \(y = 40 - 3(0) = 40 \) → (0, 40)

  • If \(x = 1\):
    \(y = 40 - 3(1) = 37 \) → (1, 37)

  • If \(x = 2\):
    \(y = 40 - 3(2) = 34 \) → (2, 34)

  • If \(x = 3\):
    \(y = 40 - 3(3) = 31 \) → (3, 31)

  • If \(x = 4\):
    \(y = 40 - 3(4) = 28 \) → (4, 28)

  • If \(x = 5\):
    \(y = 40 - 3(5) = 25 \) → (5, 25)

  • If \(x = 6\):
    \(y = 40 - 3(6) = 22 \) → (6, 22)

  • If \(x = 7\):
    \(y = 40 - 3(7) = 19 \) → (7, 19)

  • If \(x = 8\):
    \(y = 40 - 3(8) = 16 \) → (8, 16)

  • If \(x = 9\):
    \(y = 40 - 3(9) = 13 \) → (9, 13)

  • If \(x = 10\):
    \(y = 40 - 3(10) = 10 \) → (10, 10)

  • If \(x = 11\):
    \(y = 40 - 3(11) = 7 \) → (11, 7)

  • If \(x = 12\):
    \(y = 40 - 3(12) = 4 \) → (12, 4)

  • If \(x = 13\):
    \(y = 40 - 3(13) = 1 \) → (13, 1)

List of Viable Solutions

Thus, the list of viable solutions (combinations of board books \(x\) and soft cover books \(y\)) is:

  1. (0, 40)
  2. (1, 37)
  3. (2, 34)
  4. (3, 31)
  5. (4, 28)
  6. (5, 25)
  7. (6, 22)
  8. (7, 19)
  9. (8, 16)
  10. (9, 13)
  11. (10, 10)
  12. (11, 7)
  13. (12, 4)
  14. (13, 1)

These pairs represent all possible combinations of board books and soft cover books that Santiago can buy without exceeding $20.