Application of Linear Programming in Maximizing Profit in a Small Business This is my project title so l need you to do the whole project for ordinary level

• Clearly define the decision variables.
• Write down the objective function (e.g., Maximize Profit = ...).
• List all constraints as linear inequalities.
• Add non-negativity constraints.

• For each inequality constraint, treat it as an equation (e.g., change ≤ to =).
• Plot the line for each equation on a graph.
• Determine the feasible region: This is the area on the graph that satisfies all the constraints, including the non-negativity constraints (usually the first quadrant). Shade this region.

• The optimal solution (maximum or minimum) will always occur at one of the corner points (vertices) of the feasible region.
• Find the coordinates (x, y values) of each corner point. This might involve solving simultaneous equations for intersecting lines.

• Substitute the coordinates of each corner point into the objective function.
• Calculate the value of the objective function (e.g., profit) for each corner point.

• For a maximization problem, the corner point that gives the highest value for the objective function is the optimal solution.
• For a minimization problem, it's the corner point that gives the lowest value.

5. Example: A Small Bakery

Let's consider a small bakery that makes two types of cakes: Chocolate Cake and Vanilla Cake.

Problem Description:
The bakery has limited resources:

• Flour: 10 kg available per day.
• Sugar: 8 kg available per day.
• Labor: 12 hours available per day.

To make one Chocolate Cake:

• Requires 0.5 kg flour.
• Requires 0.2 kg sugar.
• Requires 1 hour of labor.
• Generates a profit of $15.

To make one Vanilla Cake:

• Requires 0.4 kg flour.
• Requires 0.3 kg sugar.
• Requires 1 hour of labor.
• Generates a profit of $12.

The bakery wants to know how many of each cake to bake daily to maximize its total profit.

Formulation:

• Let x = number of Chocolate Cakes to bake.
• Let y = number of Vanilla Cakes to bake.

• Maximize Profit (P) = 15x + 12y

• Flour Constraint: 0.5x + 0.4y ≤ 10
• Sugar Constraint: 0.2x + 0.3y ≤ 8
• Labor Constraint: 1x + 1y ≤ 12
• Non-negativity Constraints: x ≥ 0, y ≥ 0

Graphical Solution:

• Flour: 0.5x + 0.4y = 10
  • If x = 0, 0.4y = 10 ⟹ y = 25. Point (0, 25)
  • If y = 0, 0.5x = 10 ⟹ x = 20. Point (20, 0)
• Sugar: 0.2x + 0.3y = 8
  • If x = 0, 0.3y = 8 ⟹ y = 26.67. Point (0, 26.67)
  • If y = 0, 0.2x = 8 ⟹ x = 40. Point (40, 0)
• Labor: x + y = 12
  • If x = 0, y = 12. Point (0, 12)
  • If y = 0, x = 12. Point (12, 0)

• Plot these lines on a graph. Since all constraints are "less than or equal to" (≤), the feasible region will be below or to the left of each line, and within the first quadrant (due to x ≥ 0, y ≥ 0). The feasible region will be a polygon.

The corner points of the feasible region are:

• A: (0, 0)
• B: (0, 12) (Intersection of y-axis and Labor constraint)
• C: Intersection of Labor (x + y = 12) and Flour (0.5x + 0.4y = 10)
  • From x + y = 12, y = 12 - x.
  • Substitute into Flour: 0.5x + 0.4(12 - x) = 10
  • 0.5x + 4.8 - 0.4x = 10
  • 0.1x = 5.2
  • x = 52
  • y = 12 - 52 = -40. This point is outside the feasible region (y cannot be negative). This means the intersection of Labor and Flour is not a corner point of the feasible region in this specific example. Let's re-evaluate the intersections.

Let's re-check the intersections for the feasible region. The feasible region is bounded by:

• x = 0 (y-axis)
• y = 0 (x-axis)
• x + y = 12
• 0.5x + 0.4y = 10
• 0.2x + 0.3y = 8

The corner points are:

• A: (0, 0)
• B: (0, 12) (Intersection of y-axis and x + y = 12)
• C: Intersection of x + y = 12 and 0.5x + 0.4y = 10
  • From x + y = 12, y = 12 - x.
  • Substitute into 0.5x + 0.4y = 10: 0.5x + 0.4(12 - x) = 10 0.5x + 4.8 - 0.4x = 10 0.1x = 5.2 x = 52. This is too high, meaning the Labor constraint (x+y=12) is inside the region defined by the Flour constraint (0.5x + 0.4y = 10) for relevant x,y values.

Let's find the correct intersection points.
The feasible region is bounded by:

1. x = 0
2. y = 0
3. x + y = 12
4. 0.5x + 0.4y = 10
5. 0.2x + 0.3y = 8

The corner points are:

• (0, 0)
• (0, 12) (from x + y = 12)
• Intersection of x + y = 12 and 0.2x + 0.3y = 8:
  • From x + y = 12, x = 12 - y.
  • 0.2(12 - y) + 0.3y = 8
  • 2.4 - 0.2y + 0.3y = 8
  • 0.1y = 5.6
  • y = 56. This is also too high.

Let's re-examine the constraints and their intercepts.

• Flour: (20,0), (0,25)
• Sugar: (40,0), (0, 26.67)
• Labor: (12,0), (0,12)

The Labor constraint (x + y = 12) is the tightest constraint in the region near the origin.
The feasible region will be bounded by:

• x = 0
• y = 0
• x + y = 12 (This line is below the other two lines for most of the relevant range)

Let's find the intersection of x + y = 12 with the x-axis and y-axis.

• (0, 0)
• (0, 12)
• (12, 0)

Now, let's check if any other constraints cut into this triangle.

• For x + y = 12:
  • If x = 12, y = 0.
  • If y = 12, x = 0.
• For 0.5x + 0.4y = 10:
  • If x = 12, 0.5(12) + 0.4y = 10 ⟹ 6 + 0.4y = 10 ⟹ 0.4y = 4 ⟹ y = 10. Point (12, 10). This point is above the labor line.
  • If y = 12, 0.5x + 0.4(12) = 10 ⟹ 0.5x + 4.8 = 10 ⟹ 0.5x = 5.2 ⟹ x = 10.4. Point (10.4, 12). This point is above the labor line.
• For 0.2x + 0.3y = 8:
  • If x = 12, 0.2(12) + 0.3y = 8 ⟹ 2.4 + 0.3y = 8 ⟹ 0.3y = 5.6 ⟹ y = 18.67. Point (12, 18.67). This point is above the labor line.
  • If y = 12, 0.2x + 0.3(12) = 8 ⟹ 0.2x + 3.6 = 8 ⟹ 0.2x = 4.4 ⟹ x = 22. Point (22, 12). This point is above the labor line.

This means the Labor constraint (x + y ≤ 12) is the most restrictive one, and the other two constraints (Flour and Sugar) do not form part of the boundary of the feasible region in this specific example. The feasible region is simply the triangle formed by (0,0), (12,0), and (0,12).

The corner points are:

• A: (0, 0)
• B: (12, 0) (12 Chocolate Cakes, 0 Vanilla Cakes)
• C: (0, 12) (0 Chocolate Cakes, 12 Vanilla Cakes)

• At A (0, 0): P = 15(0) + 12(0) = 0
• At B (12, 0): P = 15(12) + 12(0) = 180
• At C (0, 12): P = 15(0) + 12(12) = 144

• The maximum profit is $180, which occurs at point B.

Interpretation:
To maximize profit, the bakery should produce 12 Chocolate Cakes and 0 Vanilla Cakes per day. This will yield a maximum profit of $180.

6. Conclusion

Linear programming is a powerful tool for small businesses to make optimal decisions when faced with limited resources. By clearly defining variables, objectives, and constraints, businesses can use methods like the graphical approach to identify the best production mix, maximize profits, or minimize costs. This systematic approach helps in efficient resource allocation and strategic planning, leading to better business performance.