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A confectioner sells two types of nut mixtures: the standard-mix package and the deluxe-mix package. The standard-mix contains 100 grams of cashews and 200 grams of peanuts and sells for \[tex]$1.95. The deluxe-mix contains 150 grams of cashews and 50 grams of peanuts and sells for \$[/tex]2.25. The confectioner has 15 kilograms of cashews and 20 kilograms of peanuts available. Based on past sales, the confectioner needs to have at least as many standard-mix packages as deluxe-mix packages available. How many bags of each nut mixture should she package to maximize revenue?

Note: if we let [tex]\( x \)[/tex] represent the number of standard-mix packages and [tex]\( y \)[/tex] represent the number of deluxe-mix packages sold, then we know the revenue produced by selling the packages is given by
[tex]\[ R(x, y) = 1.95x + 2.25y \][/tex]

Clearly, to maximize revenue the confectioner would simply sell as many packages as possible. However, it is not as simple as that since there are some "constraints" that must be satisfied.

Constraint 1: There are 15,000 grams of cashews available. Since the standard-mix uses 100 grams and the deluxe-mix uses 150 grams, we need to ensure that
[tex]\[ 100x + 150y \leq 15000 \Rightarrow 2x + 3y \leq 300 \][/tex]

Constraint 2: There are 20,000 grams of peanuts available. Since the standard-mix uses 200 grams and the deluxe-mix uses 50 grams, we need to ensure that
[tex]\[ 200x + 50y \leq 20000 \Rightarrow 4x + y \leq 400 \][/tex]

Constraint 3: The confectioner needs to have at least as many standard-mix packages as she has deluxe-mix packages, so we also require that
[tex]\[ x \geq y \Rightarrow y \leq x \][/tex]

Constraints 4 and 5: Since the confectioner cannot produce a negative number of packages, it must be that
[tex]\[ x \geq 0 \text{ and } y \geq 0 \][/tex]



Answer :

GinnyAnswer
To determine the optimal number of standard-mix and deluxe-mix packages to maximize revenue, we need to formulate and solve a linear programming problem. Let's summarize and define the problem in a structured manner:

### Objective Function:
Our goal is to maximize the revenue [tex]\( R \)[/tex] given by:
[tex]\[ R(x, y) = 1.95x + 2.25y \][/tex]
where:
- [tex]\( x \)[/tex] is the number of standard-mix packages,
- [tex]\( y \)[/tex] is the number of deluxe-mix packages.

### Constraints:
1. Cashews Constraint: There are 15,000 grams of cashews available. Each standard-mix uses 100 grams and each deluxe-mix uses 150 grams:
[tex]\[ 100x + 150y \leq 15000 \][/tex]
Simplifying:
[tex]\[ 2x + 3y \leq 300 \][/tex]

2. Peanuts Constraint: There are 20,000 grams of peanuts available. Each standard-mix uses 200 grams and each deluxe-mix uses 50 grams:
[tex]\[ 200x + 50y \leq 20000 \][/tex]
Simplifying:
[tex]\[ 4x + y \leq 400 \][/tex]

3. Standard-Deluxe Ratio Constraint: The number of standard-mix packages must be at least as great as the number of deluxe-mix packages:
[tex]\[ x \geq y \][/tex]
Or equivalently:
[tex]\[ y \leq x \][/tex]

4. Non-Negativity Constraints: The number of packages cannot be negative:
[tex]\[ x \geq 0 \][/tex]
[tex]\[ y \geq 0 \][/tex]

### Solution:
Given the constraints and the objective function, the problem can be solved using linear programming methods. The results are:

- Number of standard-mix packages ([tex]\( x \)[/tex]): 90
- Number of deluxe-mix packages ([tex]\( y \)[/tex]): 40

The maximum revenue achieved by this combination is:
[tex]\[ \text{Maximum Revenue} = 1.95 \cdot 90 + 2.25 \cdot 40 \][/tex]
[tex]\[ \text{Maximum Revenue} = 175.5 + 90 \][/tex]
[tex]\[ \text{Maximum Revenue} = 265.5 \][/tex]

### Conclusion:
The confectioner should package:
- 90 standard-mix packages
- 40 deluxe-mix packages

This combination will maximize the revenue, resulting in a total revenue of \$265.50.

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