Math Problem Statement
ChemLabs uses raw materials / and // to produce two domestic cleaning solutions, A and B. The daily availabilities of raw materials / and // are 150 and 145 units, respectively. One unit of solution A consumes .5 unit of raw material / and .6 unit of raw material II, and one unit of solution B uses .5 unit of raw material / and .4 unit of raw material //. The profits per unit of solutions A and B are $8 and $10, respectively. The daily demand for solution A lies between 30 and 150 units, and that for solution B between 40 and 200 units. Find the optimal production amounts of A and B.
Solution
To solve this optimization problem, we need to maximize the profit while considering the constraints on raw materials and demand. This is a linear programming (LP) problem, where:
- Objective: Maximize profit by choosing optimal amounts of solutions and to produce.
- Constraints: Availability of raw materials, demand for solutions and , and non-negativity of production amounts.
Step 1: Define the decision variables
Let:
- = number of units of solution produced.
- = number of units of solution produced.
Step 2: Write the objective function
The total profit is given by: This is what we want to maximize.
Step 3: Define the constraints
Raw material availability constraints:
- For raw material :
- For raw material :
Demand constraints:
- The demand for solution lies between 30 and 150 units:
- The demand for solution lies between 40 and 200 units:
Non-negativity constraints:
Step 4: Linear programming formulation
The problem can now be formulated as follows:
Maximize:
Subject to:
- (raw material / constraint)
- (raw material // constraint)
- (demand for A)
- (demand for B)
- , (non-negativity constraints)
Step 5: Solve using linear programming
To find the optimal solution, we can use a method such as the Simplex Method or a solver for linear programming. I will now calculate the optimal solution for this problem.The optimal production amounts for ChemLabs are:
- Produce 100 units of solution .
- Produce 200 units of solution .
This results in a maximum profit of $2800 per day.
Would you like further details or have any other questions? Here are some related questions to explore further:
- How would the solution change if the availability of raw material increased to 160 units?
- What happens to the profit if the profit per unit of solution decreases to $9?
- Can you explain the Simplex Method used in linear programming?
- What if the demand for solution decreases to a range of 30-100 units?
- How would the solution be affected if both raw materials were more expensive?
Tip: Linear programming is useful for optimizing problems with constraints; graphical methods work for two variables, but solvers like Simplex handle larger cases efficiently.
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Math Problem Analysis
Mathematical Concepts
Linear Programming
Optimization
Constraints
Formulas
Profit = 8x1 + 10x2
0.5x1 + 0.5x2 ≤ 150 (raw material / constraint)
0.6x1 + 0.4x2 ≤ 145 (raw material // constraint)
30 ≤ x1 ≤ 150 (demand for solution A)
40 ≤ x2 ≤ 200 (demand for solution B)
Theorems
Simplex Method
Linear Optimization Theorems
Suitable Grade Level
Undergraduate Level or Advanced High School
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