Simulation · Maths · Class 12

Hunt for the most profitable plan

From the lesson The corner point method in Linear Programming. Change the values and watch what happens.

The idea behind it

NCERT §12.2.2, Examples 1–2

Step 1: draw the feasible region and find all its corners, by inspection or by solving the two boundary equations that meet at each.
Step 2: evaluate Z at every corner. Call the largest value M and the smallest m.
Step 3: if the region is bounded, M is the maximum and m the minimum of Z.
Dealer: Z = 250x + 75y is 0 at O, 5000 at A(20, 0), 6250 at B(10, 50) and 4500 at C(0, 60). The maximum profit Rs 6250 comes from buying 10 tables and 50 chairs.
B is where 5x + y = 100 meets x + y = 60: subtracting gives 4x = 40, so x = 10 and y = 50.
Maximise Z = 4x + y with x + y ≤ 50, 3x + y ≤ 90, x, y ≥ 0: the corners (0, 0), (30, 0), (20, 30), (0, 50) give 0, 120, 110, 50, so the maximum is 120 at (30, 0).
Minimise Z = 200x + 500y with x + 2y ≥ 10, 3x + 4y ≤ 24, x, y ≥ 0: the region is the triangle (0, 5), (4, 3), (0, 6), with Z = 2500, 2300, 3000, so the minimum is 2300 at (4, 3).
Minimising uses exactly the same table; only the choice of the smallest entry changes.

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