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A catering business offers two sizes of baked ziti. The small ziti dish uses 1 cup of sauce, and the large ziti dish uses 2 cups of sauce and 3 cups of cheese. The business has 100 cups of sauce and cheese on hand. The profit is [tex]$6 for small dishes and $[/tex]5 for large dishes. Let [tex]\( x \)[/tex] represent the number of small dishes and [tex]\( y \)[/tex] represent the number of large dishes.

What are the constraints for the problem?

[tex]\[
\begin{aligned}
1. & \quad x + 2y \leq 100 \quad &\text{(sauce constraint)} \\
2. & \quad 3y \leq 100 \quad &\text{(cheese constraint)} \\
3. & \quad x \geq 0 \quad &\text{(non-negative dishes)} \\
4. & \quad y \geq 0 \quad &\text{(non-negative dishes)} \\
\end{aligned}
\][/tex]

Note: The incorrect constraints provided in the original text are removed for clarity.



Answer :

GinnyAnswer
To find the constraints for the problem of maximizing the profit from selling two sizes of baked ziti with the given conditions, we need to consider the resources available and the requirements for each dish type.

1. Small Ziti Dish:
- Uses 1 cup of sauce.

2. Large Ziti Dish:
- Uses 2 cups of sauce.
- Uses 3 cups of cheese.

3. Total Resources Available:
- 100 cups of sauce.
- No explicit limit provided for cheese in the problem statement.

Considering these factors, we will now write down the constraints related to the number of dishes that can be produced:

1. Non-negativity Constraints:
- The number of small dishes (x) cannot be negative: [tex]\( x \geq 0 \)[/tex].
- The number of large dishes (y) cannot be negative: [tex]\( y \geq 0 \)[/tex].

2. Sauce Constraint:
- Each small dish uses 1 cup of sauce.
- Each large dish uses 2 cups of sauce.
- The total amount of sauce used by both types of dishes must not exceed 100 cups.
Given the total sauce available, the constraint can be written as:
[tex]\[ x + 2y \leq 100 \][/tex]

Combining these observations, the constraints for the problem are:

[tex]\[ \begin{array}{l} x \geq 0, \quad (\text{number of small dishes}) \\ y \geq 0, \quad (\text{number of large dishes}) \\ x + 2y \leq 100 \quad (\text{sauce constraint}) \end{array} \][/tex]

So, summarizing, the constraints for the problem are:
[tex]\[ \begin{array}{l} x \geq 0 \\ y \geq 0 \\ x + 2y \leq 100 \end{array} \][/tex]

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