Question

A tailor stitches trousers and shirts and each is completed by two machines \(\mathrm{I}\) and II. To complete each trousers, machines I and II must work \(3 / 2 \mathrm{~h}\) and \(2 \mathrm{~h}\) respectively, and to complete each shirt, machines I and II must work \(2 \mathrm{~h}\) and \(1 \mathrm{~h}\) respectively. Machine I may not operate for more than \(12 \mathrm{~h}\) per day and machine II not more than 11 hours per day. If the profit on each trouser and each shirt is \(\mathrm{Rs} 150\) and Rs 100 respectively, then the maximum profit is (1) \(\mathrm{Rs} 900\) (2) \(\mathrm{Rs} 500\) (3) Rs 375 (4) Rs 600

Short answer

Answer: (1) Rs 900.

Step-by-step solution

Define the variables

Let x be the number of trousers made and y be the number of shirts made.

Write the constraints as inequalities

We are given the operating time constraints for each machine. We can write these constraints as linear inequalities: Machine I: \(\frac{3}{2}x + 2y \leq 12\) (Machine I may not operate for more than 12 hours per day) Machine II: \(2x + y \leq 11\) (Machine II may not operate for more than 11 hours per day) Additionally, both x and y must be non-negative. \(x \geq 0\) \(y \geq 0\)

Define the profit function

The profit function, P(x, y), can be defined as: \(P(x, y) = 150x + 100y\) where x is the number of trousers and y is the number of shirts. We are to maximize this function under the given constraints.

Solve the system of inequalities graphically

Plot the constraints on the coordinate plane: 1. \(\frac{3}{2}x + 2y \leq 12\) (Machine I) 2. \(2x + y \leq 11\) (Machine II) 3. \(x \geq 0\) 4. \(y \geq 0\) Identify the vertices of the feasible region. In this case, there are four vertices: \((0,0)\), \((0,6)\), \((4,3)\), and \((5.5,0)\).

Evaluate the profit function at the vertices

Now, we will evaluate the profit function at each of the feasible region's vertices to determine the maximum profit: 1. \(P(0,0) = 150(0) + 100(0) = 0\) 2. \(P(0,6) = 150(0) + 100(6) = 600\) 3. \(P(4,3) = 150(4) + 100(3) = 900\) 4. \(P(5.5,0) = 150(5.5) = 100(0) = 825\) From the calculations above, we can see that the maximum profit occurs at the point \((4,3)\), which corresponds to making 4 trousers and 3 shirts. The maximum profit will be Rs 900. So the correct answer is (1) Rs 900.

Key concepts

Inequality Constraints

In linear programming, the concept of inequality constraints is essential. These are mathematical expressions that define the limitations within which a problem must be solved. They appear as inequalities that restrict the values the variables can take.
In our exercise, the variables represent the number of trousers and shirts a tailor can produce. The time each machine can operate is limited, resulting in the inequalities:

• Machine I: \(\frac{3}{2}x + 2y \leq 12\)
• Machine II: \(2x + y \leq 11\)

The constraints ensure that the total operating time of each machine does not exceed their daily limits. Additionally, the non-negativity constraints \(x \geq 0\) and \(y \geq 0\) reflect that negative production of trousers or shirts is not possible. These constraints are fundamental in defining the solution space for the problem, guiding how many trousers and shirts can be optimally produced.

Graphical Method

The graphical method is a visual approach to solving linear programming problems, particularly useful for problems with two variables. It involves plotting the inequality constraints on a coordinate plane and identifying the feasible region.First, each inequality is plotted as a line on the graph. The area that satisfies all the inequalities is known as the feasible region. In this example, boundaries form from:

• \(\frac{3}{2}x + 2y = 12\)
• \(2x + y = 11\)

The feasible region is where these inequalities overlap, including the axes constraints \(x \geq 0\) and \(y \geq 0\). This region is critical because it defines the possible combinations of trousers and shirts that can be produced under the given constraints. By examining the vertices of this region, we can determine potential solutions that optimize the objective function within the constraints.

Profit Maximization

Profit maximization is the primary objective in many linear programming problems, especially those involving resource allocation like our tailoring exercise. The aim is to find the quantity of products, trousers (\(x\)) and shirts (\(y\)), that generates the highest profit while adhering to all constraints.The profit function for this exercise is:\[P(x, y) = 150x + 100y\]This formula calculates profit based on the number of trousers and shirts produced. The goal is to maximize this function within the feasible region determined by the constraints. Solving this involves calculating the profit at each vertex of the feasible region. Among the calculated profits, the highest value, in our exercise at point \((4,3)\), indicates the maximum profit achievable, which is Rs 900.

Feasible Region

The feasible region is a pivotal concept in linear programming. It represents all possible solutions that satisfy the given constraints in a problem. This region is typically a polygonal area on the graph, bounded by the lines of the inequality constraints.In our tailoring exercise, the feasible region is determined by the constraints:

• \(\frac{3}{2}x + 2y \leq 12\)
• \(2x + y \leq 11\)
• \(x \geq 0\)
• \(y \geq 0\)

The intersection points of these lines form vertices that represent possible solutions. Checking the profit at these vertices helps determine which point yields the maximum profit. Regardless of the context, the feasible region is the key area where all constraints of a problem hold true, and where optimal solutions are typically found. In graphical terms, it is the actionable space where the problem's objective can be maximized.