A candy company has 138 kg of​ chocolate-covered nuts and 78 kg of​ chocolate-covered raisins to be sold as two different mixes. One mix will contain half nuts and half raisins and will sell for​ $7 per kg. The other mix will contain three fourths
nuts and one fourth
raisins and will sell for​ $9.50 per kg. Complete parts a. and b.
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Part 1
​(a) How many kilograms of each mix should the company prepare for the maximum​ revenue? Find the maximum revenue.
The company should prepare

enter your response here kg of the first mix and

enter your response here kg of the second mix for a maximum revenue of ​$

enter your response here.

To solve this problem, we need to set up a system of equations and optimize the revenue, which is generated from selling two different mixes. Let's define the variables and proceed step by step:

### Step 1: Define variables
Let:
- $ x $ = kilograms of the first mix (half nuts and half raisins)
- $ y $ = kilograms of the second mix (three-fourths nuts and one-fourth raisins)

### Step 2: Set up the constraints based on the total amount of nuts and raisins available
From the problem, the total amount of chocolate-covered nuts is 138 kg, and the total amount of chocolate-covered raisins is 78 kg.

- The first mix is half nuts and half raisins. So, for $ x $ kg of the first mix:
  - Nuts: $ 0.5x $
  - Raisins: $ 0.5x $

- The second mix contains three-fourths nuts and one-fourth raisins. So, for $ y $ kg of the second mix:
  - Nuts: $ 0.75y $
  - Raisins: $ 0.25y $

Now, the total amounts of nuts and raisins must be used up:
1. **Nuts constraint**: $ 0.5x + 0.75y = 138 $
2. **Raisins constraint**: $ 0.5x + 0.25y = 78 $

### Step 3: Set up the revenue function
The revenue function is based on the amount of each mix produced and sold:
- The first mix sells for $7 per kg, so the revenue from the first mix is $ 7x $.
- The second mix sells for $9.50 per kg, so the revenue from the second mix is $ 9.5y $.

Thus, the total revenue $ R $ is:
\[
R = 7x + 9.5y
\]

### Step 4: Solve the system of equations
We need to solve the two equations:
1. $ 0.5x + 0.75y = 138 $
2. $ 0.5x + 0.25y = 78 $

By solving these equations, we can find $ x $ and $ y $, which will allow us to calculate the maximum revenue.

Let me calculate this.The company should prepare:
- 96 kg of the first mix (half nuts and half raisins)
- 120 kg of the second mix (three-fourths nuts and one-fourth raisins)

### Step 5: Calculate the maximum revenue
The total revenue $ R $ is given by:
\[
R = 7x + 9.5y
\]
Substitute $ x = 96 $ and $ y = 120 $ into this equation to find the maximum revenue.

Let me calculate that next.The maximum revenue the company can achieve is $1,812.

Thus, for the maximum revenue, the company should prepare:
- 96 kg of the first mix
- 120 kg of the second mix
- Maximum revenue: $1,812

Would you like more details or have any questions?

Here are some related questions to consider:
1. How would the result change if the price per kg of the second mix was $10?
2. What if the company only had 100 kg of nuts available, how would that affect the optimal solution?
3. How would we approach this problem if the goal was to minimize costs instead of maximizing revenue?
4. How do changes in ingredient proportions affect overall revenue?
5. What would happen if the company wanted to maintain a 60:40 ratio of nuts to raisins in both mixes?

**Tip:** When solving optimization problems, setting up the correct system of equations based on constraints is key to finding the optimal solution efficiently.

A candy company has 138 kg of chocolate-covered nuts and 78 kg of chocolate-covered raisins to be sold as two different mixes. One mix will contain half nuts and half raisins and will sell for $7 per kg. The other mix will contain three-fourths nuts and one-fourth raisins and will sell for $9.50 per kg. (a) How many kilograms of each mix should the company prepare for the maximum revenue? Find the maximum revenue.

Learn how to maximize revenue by solving a system of equations involving two mixes of chocolate-covered nuts and raisins. The problem includes constraints for ingredient quantities and utilizes key algebraic concepts, such as linear equations and optimization. Suitable for Grades 10-12.

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