12 min readPathway sheetOften a 3 or 5-mark question; also a 4-mark case-based question

Graphical method: the corner point method

Draw the feasible region, list its corner points, work out Z at each, and choose the largest or smallest. This is the corner point method.

Corner point methodFind every corner point, evaluate Z = ax + by at each, then compare

Bounded regionZ always has both a maximum and a minimum, each at a corner point

Two corners give the same best valueEvery point on the segment joining them is optimal (infinitely many optimal solutions)

The corner point method

How to draw the graph for an LPP answer

1
Draw the x and y-axes with a clear scale, and label them x and y.
2
Draw each constraint line using its two intercepts, and write its equation on it.
3
Shade the feasible region and name it, for example 'OABC'.
4
Mark each corner point with a dot and its coordinates.
5★
Beside the graph, draw a table: corner point and value of Z. Mark the maximum or minimum.

★ key step

Example 1: maximum on a bounded region

Problem

Solve graphically: Maximise Z = 40x + 60y subject to 2x + y ≤ 12, x + 2y ≤ 12, x ≥ 0, y ≥ 0.

Working

The feasible region is OABC with O(0, 0), A(6, 0), B(4, 4), C(0, 6); it is bounded.
Z at O = 0
Z at A = 40(6) = 240
Z at B = 40(4) + 60(4) = 160 + 240 = 400
Z at C = 60(6) = 360
The largest value is 400, at B(4, 4).

Answer

Maximum Z = 400 at x = 4, y = 4.

Example 2: three non-trivial constraints, maximum and minimum

Problem

Find the minimum and maximum of Z = 3x + 5y subject to x + y ≥ 2, x ≤ 4, y ≤ 3, x ≥ 0, y ≥ 0.

Working

Lines: x + y = 2 through (2, 0) and (0, 2); x = 4; y = 3.
O gives 0 ≥ 2, false, so shade away from O for x + y ≥ 2, and towards O for x ≤ 4 and y ≤ 3.
Corners: A(2, 0), B(4, 0), C(4, 3), D(0, 3), E(0, 2).
Z: A = 6, B = 12, C = 12 + 15 = 27, D = 15, E = 10
The smallest is 6 at A; the largest is 27 at C.

Answer

Minimum Z = 6 at (2, 0); maximum Z = 27 at (4, 3).

Example 3: more than one optimal solution

Problem

Maximise Z = 6x + 12y subject to 2x + y ≤ 12, x + 2y ≤ 12, x ≥ 0, y ≥ 0.

Working

Corners as in Example 1: O(0, 0), A(6, 0), B(4, 4), C(0, 6).
Z at O = 0, at A = 36, at B = 24 + 48 = 72, at C = 72
The maximum 72 occurs at both B and C.

Answer

Maximum Z = 72 at every point of the segment BC joining (4, 4) and (0, 6).

Common mistakes

Missing a corner: check where each pair of boundary lines meets.
Shading the wrong side: always test a point such as O.
Including a crossing point that breaks another constraint: check every corner in every constraint.
Not stating the answer in words: write 'Maximum Z = ... at x = ..., y = ...'.

In real life

A Tiruppur garment unit can use this method to decide how many shirts and kurtas to make in a week for the highest profit, given its machine hours.

Exam angle

1 marks: “The corner points of a bounded feasible region are (0, 3), (2, 0), (5, 0), (5, 4) and (0, 6). Find the maximum of Z = 3x + 2y.” Z = 6, 6, 15, 23 and 12. The maximum is 23, at (5, 4).

3 marks: “Solve graphically: Maximise Z = 5x + 4y subject to x + y ≤ 8, 2x + y ≤ 12, x ≥ 0, y ≥ 0.” The lines meet at (4, 4). Corners (0, 0), (6, 0), (4, 4), (0, 8). Z = 0, 30, 36, 32. Maximum Z = 36 at (4, 4).

5 marks: “A farmer has 10 hectares for ragi and groundnut, and 24 workers. Ragi needs 2 workers per hectare and groundnut needs 3. Profit per hectare is ₹ 6000 for ragi and ₹ 8000 for groundnut (example rates). How should she divide the land for the most profit?” Maximise Z = 6000x + 8000y subject to x + y ≤ 10, 2x + 3y ≤ 24, x ≥ 0, y ≥ 0. The lines meet at (6, 4). Corners (0, 0), (10, 0), (6, 4), (0, 8): Z = 0, 60000, 68000, 64000. Best: 6 ha of ragi and 4 ha of groundnut, profit ₹ 68,000.

Remember it

Draw, shade, corners, table, compare.