Answer :
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]
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]