Answered

A company manufactures two products, [tex]$X$[/tex] and [tex]$Y$[/tex], which require the following resources: machine capacities M1, M2, and M3. The available capacities are 50, 25, and 15 hours respectively in the planning period. Product [tex]$X$[/tex] requires 1 hour of machine M2 and 1 hour of machine M3. Product [tex]$Y$[/tex] requires 2 hours of machine M1, 2 hours of machine M2, and 1 hour of machine M3. The profit contribution of products [tex]$X$[/tex] and [tex]$Y$[/tex] are 5 and 4 respectively.

\begin{tabular}{|c|c|c|c|}
\hline
Machines & [tex]$X$[/tex] & [tex]$Y$[/tex] & Availability in hours \\
\hline
[tex]$M_1$[/tex] & 0 & 2 & 50 \\
\hline
[tex]$M_2$[/tex] & 1 & 2 & 25 \\
\hline
[tex]$M_3$[/tex] & 1 & 1 & 15 \\
\hline
Profit in Rs. Per unit & 5 & 4 & \\
\hline
\end{tabular}

Formulate the constraints and the objective function.



Answer :

GinnyAnswer
To formulate the constraints and the objective function, we need to translate the given problem into a mathematical model. We start by identifying our variables, constraints, and the objective function:

### Variables
Let:
- [tex]\(x\)[/tex] be the number of units of product [tex]\(X\)[/tex] to produce.
- [tex]\(y\)[/tex] be the number of units of product [tex]\(Y\)[/tex] to produce.

### Objective Function
The objective is to maximize the total profit. Given that the profit contribution of product [tex]\(X\)[/tex] is \[tex]$5 per unit and the profit contribution of product \(Y\) is \$[/tex]4 per unit, we can write the objective function as:
[tex]\[ \text{Maximize } Z = 5x + 4y \][/tex]

### Constraints
The constraints are based on the available machine hours and the requirements per product.

#### For Machine M1:
Product [tex]\(X\)[/tex] requires 0 hours and product [tex]\(Y\)[/tex] requires 2 hours. The total available hours are 50. Consequently, the constraint is:
[tex]\[ 2y \leq 50 \][/tex]

#### For Machine M2:
Product [tex]\(X\)[/tex] requires 1 hour and product [tex]\(Y\)[/tex] requires 2 hours. The total available hours are 25. Consequently, the constraint is:
[tex]\[ x + 2y \leq 25 \][/tex]

#### For Machine M3:
Product [tex]\(X\)[/tex] requires 1 hour and product [tex]\(Y\)[/tex] requires 1 hour. The total available hours are 15. Consequently, the constraint is:
[tex]\[ x + y \leq 15 \][/tex]

### Non-negativity Constraints
We also need to ensure that the number of units produced cannot be negative:
[tex]\[ x \geq 0 \][/tex]
[tex]\[ y \geq 0 \][/tex]

### Formulated Problem
To summarize, the linear programming problem can be formulated as follows:

Objective Function:
[tex]\[ \text{Maximize } Z = 5x + 4y \][/tex]

Subject to the Constraints:
[tex]\[ 2y \leq 50 \][/tex]
[tex]\[ x + 2y \leq 25 \][/tex]
[tex]\[ x + y \leq 15 \][/tex]
[tex]\[ x \geq 0 \][/tex]
[tex]\[ y \geq 0 \][/tex]

### Optimal Solution
The solution to this linear programming problem yields the optimal number of products [tex]\(X\)[/tex] and [tex]\(Y\)[/tex] to produce, as well as the maximum profit. Based on the calculations, the optimal solution is:

- [tex]\( x = 15 \)[/tex]
- [tex]\( y = 0 \)[/tex]
- Maximum Profit [tex]\( Z = 75 \)[/tex]

So, it is optimal to produce 15 units of product [tex]\(X\)[/tex] and 0 units of product [tex]\(Y\)[/tex], yielding a maximum profit of \$75.