Application-Based Linear Programming Problems

Read the context and identify the two quantities to be decided. Create a table for profit or cost and resource use. Write the objective function. Convert each resource or requirement statement into a constraint. Add non-negativity conditions. Solve graphically and interpret the optimum in words with units.

The application itself may not need a special diagram, but the solved LPP normally requires a coordinate graph with feasible region and corner points when optimum is requested.

Leaving the final answer as only Z = 1300 without stating the required values of x and y loses the decision part of the problem.

Application word problems use linear programming to model real situations such as diet planning, manufacturing, transport, and allocation of limited resources.

Application word problems use linear programming to model real situations such as diet planning, manufacturing, transport, and allocation of limited resources. Application modelling pattern: decision variables x, y; objective function Z = cost/profit expression; constraints from resource limits or minimum requirements; non-negativity x >= 0, y >= 0. Use <= for limited availability and >= for required minimum intake, demand, or production. In a manufacturing problem, x and y may represent numbers of two products. Machine-hour limits, material limits, and demand conditions form constraints, while profit or cost forms the objective function.

These problems require translating a context into variables, objective function, and constraints. The mathematical solution must be interpreted in the original context. Since Class 12 LPP is graphical, problems usually involve two decision variables and linear restrictions. Application modelling pattern: decision variables x, y; objective function Z = cost/profit expression; constraints from resource limits or minimum requirements; non-negativity x >= 0, y >= 0. Use <= for limited availability and >= for required minimum intake, demand, or production. A small unit makes x standard bags and y deluxe bags. Profit is Rs 40 and Rs 60 respectively. Cutting time available is 80 hours, with each standard bag needing 2 hours and each deluxe bag needing 4 hours. Stitching time available is 60 hours, with each needing 3 hours and 2 hours respectively. Formulate and solve. Maximize Z = 40x + 60y subject to 2x + 4y <= 80, 3x + 2y <= 60, x >= 0, y >= 0. Simplify first constraint: x + 2y <= 40. Corner points are (0,0), (20,0), intersection of x + 2y = 40 and 3x + 2y = 60, and (0,20). For intersection, subtract: (3x + 2y) - (x + 2y) = 60 - 40, so 2x = 20, x = 10. Then 10 + 2y = 40, so y = 15. Evaluate Z: Z(0,0)=0, Z(20,0)=800, Z(10,15)=400+900=1300, Z(0,20)=1200. Maximum profit is Rs 1300 at x = 10, y = 15. Usually appears as a long-answer or case-study problem where students must form the LPP, draw the feasible region, evaluate corner points, and write the final real-world decision. Leaving the final answer as only Z = 1300 without stating the required values of x and y loses the decision part of the problem.