Math Problem Statement

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?

Solution

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 xx represent the number of snack-size bags produced.
  • Let yy represent the number of family-size bags produced.

Objective Function:

The company's goal is to maximize profit, which can be described as: Profit=12x+18y\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+y2400x + y \leq 2400
  2. At least 600 units of snack-size bags must be produced: x600x \geq 600
  3. At least 900 units of family-size bags must be produced: y900y \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)(x, y) = (600, 1800) (Substituting x=600x = 600 in x+y=2400x + y = 2400)
  2. (x,y)=(600,900)(x, y) = (600, 900) (This is a boundary point where both constraints are at their minimum)
  3. (x,y)=(1500,900)(x, y) = (1500, 900) (Substituting y=900y = 900 in x+y=2400x + y = 2400)

Profit at Each Point:

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

Conclusion:

The maximum profit occurs at the point (600,1800)(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.

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Math Problem Analysis

Mathematical Concepts

Linear Programming
Inequalities
Optimization

Formulas

Objective function: Profit = 12x + 18y
Constraints: x + y ≤ 2400, x ≥ 600, y ≥ 900

Theorems

Corner Point Theorem
Graphical Method in Linear Programming

Suitable Grade Level

Grades 9-12