A potato chip company makes chips to fill snack-size bags and family-size bags. <in one week, production cannot exceed 2400 units, of which at least 600 units must be for snacks-size bags and at least 900 units must be for family-size bags is $12, and the profit on a unit of family-size bags is $18. How much of each type of bag must be produced to miximize profits?

This is a linear programming problem where the objective is to maximize profit based on certain constraints. Let's define the variables and construct the necessary equations:

### Variables:
- Let $ x $ represent the number of snack-size bags produced.
- Let $ y $ represent the number of family-size bags produced.

### Objective Function:
The company's goal is to maximize profit, which can be described as:
\[
\text{Profit} = 12x + 18y
\]
where $12 is the profit per snack-size bag and $18 is the profit per family-size bag.

### Constraints:
1. The total production cannot exceed 2400 units, so:
   \[
   x + y \leq 2400
   \]
2. At least 600 units of snack-size bags must be produced:
   \[
   x \geq 600
   \]
3. At least 900 units of family-size bags must be produced:
   \[
   y \geq 900
   \]

Now, we can maximize the profit subject to these constraints using graphical or algebraic methods. First, let's graph the inequalities to find the feasible region and then find which corner of that region gives the maximum profit.

We will calculate the profit at the corner points of the feasible region formed by these constraints.

### Corner Points:
1. $ (x, y) = (600, 1800) $  (Substituting $x = 600$ in $x + y = 2400$)
2. $ (x, y) = (600, 900) $   (This is a boundary point where both constraints are at their minimum)
3. $ (x, y) = (1500, 900) $  (Substituting $y = 900$ in $x + y = 2400$)

### Profit at Each Point:
1. At $ (600, 1800) $:
   \[
   \text{Profit} = 12(600) + 18(1800) = 7200 + 32400 = 39600
   \]
2. At $ (600, 900) $:
   \[
   \text{Profit} = 12(600) + 18(900) = 7200 + 16200 = 23400
   \]
3. At $ (1500, 900) $:
   \[
   \text{Profit} = 12(1500) + 18(900) = 18000 + 16200 = 34200
   \]

### Conclusion:
The maximum profit occurs at the point $ (600, 1800) $, where the company should produce **600 snack-size bags** and **1800 family-size bags** to achieve the maximum profit of **$39,600**.

Would you like further clarification on any steps?

---

Here are 5 follow-up questions for deeper understanding:
1. How would the solution change if the profit per family-size bag increased?
2. What would happen to the feasible region if the total production limit were higher?
3. How could you solve this problem using an algebraic method like the Simplex algorithm?
4. What if the company required more snack-size bags than family-size bags?
5. Could external factors like demand influence the optimal solution?

**Tip**: Linear programming problems often have their optimal solutions at the vertices (corners) of the feasible region.

A potato chip company makes chips to fill snack-size bags and family-size bags. In one week, production cannot exceed 2400 units, of which at least 600 units must be for snack-size bags and at least 900 units must be for family-size bags. The profit on a unit of snack-size bags is $12, and the profit on a unit of family-size bags is $18. How much of each type of bag must be produced to maximize profits?

Explore how to maximize profits for a potato chip company producing snack-size and family-size bags with constraints on production. Learn how to apply linear programming, inequalities, and the corner point method. Suitable for Grades 9-12.

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