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An optimisation problem asks for the maximum (profit) or minimum (cost) of a quantity. A linear programming problem (LPP) maximises or minimises a linear objective function of the decision variables, subject to linear constraints and the non-negative restrictions.
| Term | Meaning |
|---|---|
| Objective function | The linear function Z = ax + by to be maximised or minimised |
| Decision variables | x and y, the quantities we choose (non-negative) |
| Constraints | The linear inequalities or equations the variables must satisfy |
| Non-negative restrictions | x ≥ 0, y ≥ 0 |
| Feasible region | The common region of all the constraints, x, y ≥ 0 included |
| Feasible solution | A point in the feasible region (inside or on its boundary) |
| Infeasible solution | A point outside the feasible region |
| Optimal solution | A feasible point that gives the maximum or minimum value of Z |
A dealer has Rs 50,000 and storage for at most 60 pieces. A table costs Rs 2500 and gives profit Rs 250. A chair costs Rs 500 and gives profit Rs 75. Let x tables and y chairs be bought.
The corner points of OABC are O(0, 0), A(20, 0), B(10, 50), C(0, 60).
| Corner point | Z = 250x + 75y |
|---|---|
| (0, 0) | 0 |
| (20, 0) | 5000 |
| (10, 50) | 6250 |
| (0, 60) | 4500 |
The maximum profit is Rs 6250, from 10 tables and 50 chairs.
Maximise Z = 4x + y subject to x + y ≤ 50, 3x + y ≤ 90, x, y ≥ 0.
| Corner point | Z = 4x + y |
|---|---|
| (0, 0) | 0 |
| (30, 0) | 120 |
| (20, 30) | 110 |
| (0, 50) | 50 |
The region is bounded, so the maximum is 120 at (30, 0). The corner (20, 30) comes from solving x + y = 50 and 3x + y = 90.
The region can be bounded, unbounded or have no points at all (then there is no feasible solution).
Minimise and maximise Z = 3x + 9y subject to x + 3y ≤ 60, x + y ≥ 10, x ≤ y, x, y ≥ 0.
| Corner point | Z = 3x + 9y |
|---|---|
| (0, 10) | 90 |
| (5, 5) | 60 |
| (15, 15) | 180 |
| (0, 20) | 180 |
The minimum is 60 at B(5, 5). The maximum 180 occurs at both C and D, so it occurs at every point of the segment CD.
Minimise Z = −50x + 20y subject to 2x − y ≥ −5, 3x + y ≥ 3, 2x − 3y ≤ 12, x, y ≥ 0.
| Corner point | Z = −50x + 20y |
|---|---|
| (1, 0) | −50 |
| (6, 0) | −300 |
| (0, 5) | 100 |
| (0, 3) | 60 |
The smallest corner value is −300 at (6, 0), but the region is unbounded and the open half plane −50x + 20y < −300 meets it, so Z has no minimum. (The maximum 100 at (0, 5) does exist, since the half plane −50x + 20y > 100 misses the region.)
Minimise Z = 3x + 2y subject to x + y ≥ 8, 3x + 5y ≤ 15, x, y ≥ 0. For x + y ≥ 8 we need 3x + 5y ≥ 24 > 15, so no point satisfies both. There is no feasible solution.
| Situation | Conclusion |
|---|---|
| Bounded region | Max = largest corner value, min = smallest |
| Unbounded region | Test the open half plane before accepting M or m |
| Two corners tie for the optimum | Every point of the segment between them is optimal |
| Constraints contradict each other | No feasible solution |
| Constraint ≥ | Shade the side away from the origin if the origin fails it |
Try each problem first, then open the solution to check your working.
For the constraints 2x + y ≤ 10, x + 3y ≤ 15, x ≥ 0, y ≥ 0, decide which of the points (2, 3), (4, 3), (5, 0), (3, 5) are feasible solutions.
(2, 3): 4 + 3 = 7 ≤ 10 and 2 + 9 = 11 ≤ 15. Feasible.
(4, 3): 8 + 3 = 11 > 10. Not feasible.
(5, 0): 10 ≤ 10 and 5 ≤ 15. Feasible.
(3, 5): 6 + 5 = 11 > 10. Not feasible.
Answer: (2, 3) and (5, 0) are feasible; (4, 3) and (3, 5) are not
A shop sells notebooks (profit Rs 8 each, cost Rs 20) and pens (profit Rs 3 each, cost Rs 5). With Rs 600 to spend and room for at most 60 items, write the LPP for maximum profit. Let x notebooks and y pens be bought.
Cost: 20x + 5y ≤ 600, that is 4x + y ≤ 120.
Space: x + y ≤ 60.
Non-negativity: x ≥ 0, y ≥ 0. Objective: Z = 8x + 3y.
Answer: Maximise Z = 8x + 3y subject to 4x + y ≤ 120, x + y ≤ 60, x ≥ 0, y ≥ 0
Solve the problem of the previous question.
Corner points of the (bounded) feasible region and the value of Z at each:
| Corner point | Z = 8x + 3y |
|---|---|
| (0, 0) | 0 |
| (30, 0) | 240 |
| (20, 40) | 280 |
| (0, 60) | 180 |
The largest value is the maximum.
Answer: Maximum Z = 280 at (20, 40), i.e. 20 notebooks and 40 pens
A diet needs at least 12 units of vitamin A and 16 units of vitamin B. Food P costs Rs 4 per unit and gives 3 units of A and 2 of B. Food Q costs Rs 5 per unit and gives 2 units of A and 4 of B. Find the cheapest diet.
Let x units of P and y units of Q. Minimise Z = 4x + 5y subject to 3x + 2y ≥ 12, 2x + 4y ≥ 16, x, y ≥ 0.
The corner points are:
| Corner point | Z = 4x + 5y |
|---|---|
| (8, 0) | 32 |
| (0, 6) | 30 |
| (2, 3) | 23 |
The region is unbounded, so check that no feasible point has Z < 23:
4x + 5y = (3/4)(3x + 2y) + (7/8)(2x + 4y) ≥ (3/4)(12) + (7/8)(16) = 23. So the minimum is 23.
Answer: Minimum Z = 23 at (2, 3), i.e. 2 units of P and 3 units of Q for Rs 23
Product P needs 2 hours on machine A and 1 hour on machine B, and earns Rs 30. Product Q needs 1 hour on A and 2 hours on B, and earns Rs 40. Machine A is available for 100 hours and B for 80 hours. Find the maximum profit.
Maximise Z = 30x + 40y subject to 2x + y ≤ 100, x + 2y ≤ 80, x, y ≥ 0.
The corner (40, 20) is where 2x + y = 100 meets x + 2y = 80.
| Corner point | Z = 30x + 40y |
|---|---|
| (0, 0) | 0 |
| (50, 0) | 1500 |
| (40, 20) | 2000 |
| (0, 40) | 1600 |
Answer: Maximum Z = 2000 at (40, 20), i.e. 40 of P and 20 of Q
Maximise Z = 2x + 3y subject to x + y ≤ 6, x + 3y ≤ 12, x, y ≥ 0.
Corner points and values (the corner (3, 3) is where x + y = 6 meets x + 3y = 12):
| Corner point | Z = 2x + 3y |
|---|---|
| (0, 0) | 0 |
| (6, 0) | 12 |
| (3, 3) | 15 |
| (0, 4) | 12 |
The region is bounded.
Answer: Maximum Z = 15 at (3, 3)
Minimise Z = 5x + 4y subject to x + y ≥ 4, x ≤ 5, y ≤ 5, x, y ≥ 0.
The region is bounded by x + y ≥ 4, x ≤ 5, y ≤ 5 and the axes.
| Corner point | Z = 5x + 4y |
|---|---|
| (4, 0) | 20 |
| (5, 0) | 25 |
| (5, 5) | 45 |
| (0, 5) | 20 |
| (0, 4) | 16 |
Answer: Minimum Z = 16 at (0, 4)
Maximise Z = 7x + 5y subject to 2x + y ≤ 10, x + 3y ≤ 15, x, y ≥ 0.
The corner (3, 4) is where 2x + y = 10 meets x + 3y = 15: y = 10 − 2x gives x + 30 − 6x = 15, so x = 3 and y = 4.
| Corner point | Z = 7x + 5y |
|---|---|
| (0, 0) | 0 |
| (5, 0) | 35 |
| (3, 4) | 41 |
| (0, 5) | 25 |
Answer: Maximum Z = 41 at (3, 4)
Minimise Z = x + 2y subject to x + y ≥ 3, x + y ≤ 7, x ≤ 5, x, y ≥ 0.
The region lies between the lines x + y = 3 and x + y = 7 and is bounded.
| Corner point | Z = x + 2y |
|---|---|
| (3, 0) | 3 |
| (5, 0) | 5 |
| (5, 2) | 9 |
| (0, 7) | 14 |
| (0, 3) | 6 |
Answer: Minimum Z = 3 at (3, 0)
A firm makes two products. Maximise Z = 40x + 30y subject to x + y ≤ 12, 2x + y ≤ 16, x, y ≥ 0.
The corner (4, 8) is where x + y = 12 meets 2x + y = 16.
| Corner point | Z = 40x + 30y |
|---|---|
| (0, 0) | 0 |
| (8, 0) | 320 |
| (4, 8) | 400 |
| (0, 12) | 360 |
Answer: Maximum Z = 400 at (4, 8)
Minimise Z = 2x + 3y subject to x + y ≥ 5, x + 3y ≥ 9, x, y ≥ 0.
The region is unbounded. Corner points and values:
| Corner point | Z = 2x + 3y |
|---|---|
| (3, 2) | 12 |
| (9, 0) | 18 |
| (0, 5) | 15 |
To confirm 12 is the minimum, show that Z < 12 has no feasible point:
2x + 3y = (3/2)(x + y) + (1/2)(x + 3y) ≥ (3/2)(5) + (1/2)(9) = 12.
Answer: Minimum Z = 12 at (3, 2)
Maximise Z = 3x + 2y subject to x + y ≥ 4, x ≥ 1, x, y ≥ 0.
The region is unbounded and extends to the right and upward.
For any large t the point (t, 0) with t ≥ 4 is feasible and Z = 3t, which exceeds every bound.
So the open half plane 3x + 2y > M meets the region for every M.
Answer: Z has no maximum value
A baker makes x cakes (profit Rs 20) and y pastries (profit Rs 10). Flour allows 2x + y ≤ 60 and labour allows x + y ≤ 40. Maximise the profit Z = 20x + 10y.
| Corner point | Z = 20x + 10y |
|---|---|
| (0, 0) | 0 |
| (30, 0) | 600 |
| (20, 20) | 600 |
| (0, 40) | 400 |
The maximum 600 occurs at the corners (30, 0) and (20, 20). The line 20x + 10y = 600 is the same as 2x + y = 60, so every point of the segment joining them gives Z = 600.
Answer: Maximum Z = 600, attained at (30, 0), (20, 20) and every point between them
Show that the problem of maximising Z = x + 2y subject to x + 2y ≤ 4, x + y ≥ 6, x, y ≥ 0 has no solution.
Since y ≥ 0, x + y ≤ x + 2y ≤ 4. But x + y ≥ 6 is required, and 4 < 6.
No point satisfies both, so the feasible region is empty.
Answer: No feasible region, so no solution
Minimise Z = −x − y subject to x + y ≥ 2, x − y ≤ 4, x, y ≥ 0.
Points (t, t) with t ≥ 1 satisfy x + y = 2t ≥ 2 and x − y = 0 ≤ 4, so they are feasible.
There Z = −2t, which falls without limit as t grows.
Answer: Z has no minimum value
A person invests at most Rs 50,000 (in thousands: x + y ≤ 50) in two schemes. At least Rs 10,000 must go into scheme A (8% a year) and at most Rs 30,000 into scheme B (10% a year). Taking x and y in thousands of rupees, the yearly return is Z = 80x + 100y rupees. Find the investment with the largest return.
Maximise Z = 80x + 100y (x, y in thousands of rupees) subject to x + y ≤ 50, x ≥ 10, y ≤ 30, x, y ≥ 0.
| Corner point | Z = 80x + 100y |
|---|---|
| (10, 0) | 800 |
| (50, 0) | 4000 |
| (20, 30) | 4600 |
| (10, 30) | 3800 |
Answer: Maximum Z = 4600 at (20, 30), i.e. Rs 20,000 in A and Rs 30,000 in B for a return of Rs 4600
A factory makes x units of P (profit Rs 15) and y units of Q (profit Rs 10). Machine time gives 2x + y ≤ 40, labour gives x + y ≤ 30, and the market takes at most 15 units of P. Find the maximum profit.
Maximise Z = 15x + 10y subject to 2x + y ≤ 40, x + y ≤ 30, x ≤ 15, x, y ≥ 0.
The corner (10, 20) is where 2x + y = 40 meets x + y = 30; (15, 10) is where x = 15 meets 2x + y = 40.
| Corner point | Z = 15x + 10y |
|---|---|
| (0, 0) | 0 |
| (15, 0) | 225 |
| (15, 10) | 325 |
| (10, 20) | 350 |
| (0, 30) | 300 |
Answer: Maximum Z = 350 at (10, 20)
A mix uses x kg of A (Rs 3 per kg) and y kg of B (Rs 2 per kg). The mix must weigh at least 10 kg, with at most 8 kg of A and at most 8 kg of B. Find the least cost.
Minimise Z = 3x + 2y subject to x + y ≥ 10, x ≤ 8, y ≤ 8, x, y ≥ 0.
| Corner point | Z = 3x + 2y |
|---|---|
| (8, 2) | 28 |
| (8, 8) | 40 |
| (2, 8) | 22 |
Answer: Minimum Z = 22 at (2, 8)
In the shop problem (maximise Z = 8x + 3y with 4x + y ≤ 120) the space limit is reduced from 60 to 45 items. Find the new maximum profit.
The constraint becomes x + y ≤ 45. The corner (25, 20) is where 4x + y = 120 meets x + y = 45.
| Corner point | Z = 8x + 3y |
|---|---|
| (0, 0) | 0 |
| (30, 0) | 240 |
| (25, 20) | 260 |
| (0, 45) | 135 |
Answer: Maximum Z = 260 at (25, 20) (down from Rs 280)
Maximise Z = 6x + 4y subject to 3x + 2y ≤ 12, x ≤ 3, x, y ≥ 0.
| Corner point | Z = 6x + 4y |
|---|---|
| (0, 0) | 0 |
| (3, 0) | 18 |
| (3, 3/2) | 24 |
| (0, 6) | 24 |
Two corners, (3, 3/2) and (0, 6), give the same maximum. The objective lines 6x + 4y = c are parallel to the boundary 3x + 2y = 12, so the maximum occurs along the whole edge between them.
Answer: Maximum Z = 24 at (3, 3/2), (0, 6) and every point between them
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Draw the lines for the constraints and shade the feasible region. Its corner points are:
| Corner point | Z = 3x + 4y |
|---|---|
| (0, 0) | 0 |
| (4, 0) | 12 |
| (0, 4) | 16 |
The region is the triangle OAB, which is bounded, so the maximum occurs at a corner point.
Answer: Maximum Z = 16 at (0, 4)
Draw the lines for the constraints and shade the feasible region. Its corner points are:
| Corner point | Z = −3x + 4y |
|---|---|
| (0, 0) | 0 |
| (4, 0) | −12 |
| (2, 3) | 6 |
| (0, 4) | 16 |
The corner (2, 3) is the intersection of x + 2y = 8 and 3x + 2y = 12.
The region is bounded, so the smallest value is the minimum.
Answer: Minimum Z = −12 at (4, 0)
Draw the lines for the constraints and shade the feasible region. Its corner points are:
| Corner point | Z = 5x + 3y |
|---|---|
| (0, 0) | 0 |
| (2, 0) | 10 |
| (20/19, 45/19) | 235/19 |
| (0, 3) | 9 |
The corner (20/19, 45/19) is the intersection of 3x + 5y = 15 and 5x + 2y = 10: from the two equations x = 20/19 and y = 45/19.
Answer: Maximum Z = 235/19 at (20/19, 45/19)
Draw the lines for the constraints and shade the feasible region. Its corner points are:
| Corner point | Z = 3x + 5y |
|---|---|
| (3/2, 1/2) | 7 |
| (3, 0) | 9 |
| (0, 2) | 10 |
The feasible region is unbounded, so the smallest value 7 is the minimum only if the open half plane 3x + 5y < 7 has no point in common with the region.
Note that 3x + 5y = (x + 3y) + 2(x + y) ≥ 3 + 2(2) = 7 for every feasible point, so no feasible point has Z < 7.
Answer: Minimum Z = 7 at (3/2, 1/2)
Draw the lines for the constraints and shade the feasible region. Its corner points are:
| Corner point | Z = 3x + 2y |
|---|---|
| (0, 0) | 0 |
| (5, 0) | 15 |
| (4, 3) | 18 |
| (0, 5) | 10 |
The corner (4, 3) is the intersection of x + 2y = 10 and 3x + y = 15.
Answer: Maximum Z = 18 at (4, 3)
Draw the lines for the constraints and shade the feasible region. Its corner points are:
| Corner point | Z = x + 2y |
|---|---|
| (6, 0) | 6 |
| (0, 3) | 6 |
The region is unbounded. The open half plane x + 2y < 6 has no point in common with the region, because the constraint x + 2y ≥ 6 holds throughout it. So the minimum is 6.
Z = 6 at both (0, 3) and (6, 0), so it equals 6 at every point of the segment joining them (all on the line x + 2y = 6). Hence the minimum occurs at more than two points (infinitely many).
Answer: Minimum Z = 6, attained at every point of the segment from (0, 3) to (6, 0)
Draw the lines for the constraints and shade the feasible region. Its corner points are:
| Corner point | Z = 5x + 10y |
|---|---|
| (60, 0) | 300 |
| (120, 0) | 600 |
| (60, 30) | 600 |
| (40, 20) | 400 |
The region is the bounded quadrilateral with these corners.
The maximum 600 occurs at two corners, (120, 0) and (60, 30), so it occurs at every point of the segment joining them.
Answer: Minimum Z = 300 at (60, 0). Maximum Z = 600 at (120, 0), (60, 30) and every point between them
Draw the lines for the constraints and shade the feasible region. Its corner points are:
| Corner point | Z = x + 2y |
|---|---|
| (0, 50) | 100 |
| (20, 40) | 100 |
| (50, 100) | 250 |
| (0, 200) | 400 |
The region is bounded. The minimum 100 occurs at two corners, (0, 50) and (20, 40), hence along the segment joining them.
Answer: Minimum Z = 100 at (0, 50), (20, 40) and every point between them. Maximum Z = 400 at (0, 200)
Draw the lines for the constraints and shade the feasible region. Its corner points are:
| Corner point | Z = −x + 2y |
|---|---|
| (6, 0) | −6 |
| (3, 2) | 1 |
| (4, 1) | −2 |
The region is unbounded. The largest corner value is 1 at (3, 2), but the points (3, y) with y > 2 are feasible and give Z = −3 + 2y > 1.
So the open half plane −x + 2y > 1 meets the region and Z has no maximum value.
Answer: Z has no maximum value (the region is unbounded and Z increases without limit)
The constraints say y ≥ x + 1 and y ≤ x. No point can satisfy both, because x + 1 > x.
There is no common region, so the problem has no feasible solution.
Answer: No feasible region, so no maximum exists