# A confectioner sells two types of nut mixtures. The standard-mixture package contains 100 g of cashews and 200 g of peanuts and sells for $1.85. The deluxe-mixture package contains 150 g of cashews and 50 g of peanuts and sells for$2.45. The confectioner has 15 kg of cashews and 20 kg of peanuts available. On the basis of past sales, the confectioner needs to have at least as many standard as deluxe packages available. How many bags of each mixture should he package to maximize his revenue?

## Let $x$ be the number of standard-mixture packages and $y$ be the number of deluxe-mixture packages.

From the problem description we have the following constraints:

Cashews:
$100x + 150y \leq 15000$ (1)

Peanuts:
$200x + 50y \leq 20000$ (2)

Number of standard packages should be at least as many as deluxe packages:
$x \geq y$

The revenue to be maximized is given by:
$R = 1.85x + 2.45y$ (3)

We will now solve the constraints:

1) $100x + 150y \leq 15000$

Dividing both sides by $50$ we get:

$2x + 3y \leq 300$

2) $200x + 50y \leq 20000$

Dividing both sides by $50$ we get:

$4x + y \leq 400$

Now we plot the inequality constraints graphically and find the boundary point on which the function $R(x, y) = 1.85x + 2.45y$ is maximized.

The inequality constraints plot looks like the following:

y
^
|
400|
|
+-------x.

and the relevent boundary points are $A(100, 100), \ B(150, 0)$.

Now we plug in these boundary points into the revenue function.

$R_A(100, 100) = 1.85 \times 100 + 2.45 \times 100 = 430$

$R_B(150, 0) = 1.85 \times 150 + 2.45 \times 0 = 277.5$

Since the revenue at point A is higher, the confectioner should package 100 standard-mixture packages and 100 deluxe-mixture packages to maximize his revenue.

## To maximize revenue, we need to find the number of standard-mixture packages and deluxe-mixture packages the confectioner should package. Let's call the number of standard-mixture packages x and the number of deluxe-mixture packages y.

1. Set up the objective function:
The revenue is calculated by multiplying the number of packages by the price. The objective function is:
Revenue = 1.85x + 2.45y

2. Set up the constraints:
- Weight constraint: The total weight of cashews used cannot exceed the available supply of cashews and the total weight of peanuts used cannot exceed the available supply of peanuts.
100x + 150y ≤ 15,000 (cashews weight constraint)
200x + 50y ≤ 20,000 (peanuts weight constraint)

- Minimum requirement constraint: The confectioner needs at least as many standard as deluxe packages available.
x ≥ y (minimum requirement constraint)

- Non-negative constraint: Both x and y must be non-negative.
x, y ≥ 0

3. Solve the linear programming problem:
Using a linear programming solver, input the objective function and constraints. The solver will give the optimal values for x and y that maximize the revenue.

Note: In this case, the solver will provide a solution with decimal values for x and y. However, since the number of packages should be whole numbers, round down the decimal values to the nearest whole number.

The optimal values for x and y will give the number of bags of each mixture the confectioner should package to maximize revenue.

## To maximize revenue, we need to determine the number of standard-mixture and deluxe-mixture bags that the confectioner should package. Let's denote the number of standard-mixture bags as 'x' and the number of deluxe-mixture bags as 'y'.

First, let's set up the constraints based on the available quantities of cashews and peanuts:
- Cashew constraint: 100g in standard-mixture * x + 150g in deluxe-mixture * y ≤ 15kg (15000g)
- Peanut constraint: 200g in standard-mixture * x + 50g in deluxe-mixture * y ≤ 20kg (20000g)

We also know that the confectioner needs to produce at least as many standard-mixture bags as deluxe-mixture bags:
- Standard constraint: x ≥ y

Next, let's set up the revenue equation based on the prices of the packages:
- Revenue equation: Revenue = 1.85 * x + 2.45 * y

To solve this problem, we can use linear programming techniques, such as the simplex method or graphically on a feasible region graph.

I will now solve this problem using the simplex method.

Step 1: Convert the constraints into standard form by dividing each equation by the appropriate constant(s) to make the right side equal to 1.
- Cashew constraint: (100/15000)x + (150/15000)y ≤ 1
- Peanut constraint: (200/20000)x + (50/20000)y ≤ 1
- Standard constraint: x - y ≥ 0

Step 2: Rewrite the objective function (revenue equation) by subtracting X from both sides and convert it into the standard form.
- Revenue equation: -1.85x - 2.45y ≤ 0

Step 3: Set up the initial simplex tableau:
| x | y | s1 | s2 | RHS |
| --- | --- | ---- | ---- | ----- |
| 1 | 0 | 1/15000 | 0 | 1 |
| 0 | 1 | 0 | 1/20000 | 1 |
| -1.85 | -2.45 | 0 | 0 | 0 |

Step 4: Perform the simplex algorithm to find the optimal solution.

Iterating through the simplex algorithm:

For each iteration, we select the column with the most negative value in the bottom row (excluding the RHS column).

Step 5: Pivot
- Pivot on the (1, 1) element (1/15000) since it has the most negative value in the bottom row.
- Perform row operations to make all other elements in column 1 zero.

After performing row operations:
| x | y | s1 | s2 | RHS |
| --- | --- | ---- | ---- | ----- |
| 0 | 0 | 1 | -2/5 | 1/5 |
| 1 | 0 | -1/15000 | 1/15000 | 1/15000 |
| 0 | 1 | 1/20000 | 1/20000 | 1/20000 |

The current solution is now x = 1/15000, y = 1/20000, and the maximum revenue is (1.85)(1/15000) = 0.0001233 + (2.45)(1/20000) = 0.0001225 = $0.0002458. This means that approximately 0.0001233 standard-mixture bags and 0.0001225 deluxe-mixture bags should be packaged to maximize revenue. The revenue generated would be approximately$0.0002458.

Note: The answer has been rounded to the nearest decimal places for simplicity.