Linear Programming — Practice

Question 1

Maximise Z = 4x + 3y subject to x + y <= 8, x >= 0, y >= 0.

Accepted Answer

The feasible region is the triangle with corners (0,0), (8,0) and (0,8), which is bounded, so the largest corner value is the maximum. Answer: Maximum Z = 32 at (8, 0).

Question 2

Determine whether the feasible region of x + y <= 4, x >= 0, y >= 0 is bounded or unbounded, and name its corner points.

Accepted Answer

Every constraint caps the variables from above, so the region is enclosed by a triangle and can be fitted inside a circle. Answer: Bounded; corner points (0,0), (4,0), (0,4).

Question 3

Minimise Z = 200x + 500y subject to x + 2y >= 10, 3x + 4y <= 24, x >= 0, y >= 0.

Accepted Answer

Solving the two lines gives the corner (4,3), and the other corners are (0,5) and (0,6); evaluating Z at each gives 2300, 2500 and 3000. Answer: Minimum Z = 2300 at (4, 3).

Question 4

Minimise Z = 3x + 2y subject to x + y >= 8, 3x + 5y >= 15, x >= 0, y >= 0, and justify that your answer really is the minimum.

Accepted Answer

The region is unbounded, so after finding the smallest corner value you must check that 3x + 2y < m misses the region entirely, which it does. Answer: Minimum Z = 16 at (0, 8).

Question 5

A dealer has Rs 50,000 and space for 60 pieces. A table costs Rs 2500 with profit Rs 250; a chair costs Rs 500 with profit Rs 75. Formulate this as an LPP.

Accepted Answer

Let x be the number of tables and y the number of chairs; the money and space limits each give a <= inequality, and quantities cannot be negative. Answer: Maximise Z = 250x + 75y subject to 5x + y <= 100, x + y <= 60, x >= 0, y >= 0.

Question 6

Minimise Z = 2x + 4y subject to x + 2y >= 10, x >= 0, y >= 0, and state where the minimum is attained.

Accepted Answer

Since Z = 2(x + 2y) and the constraint forces x + 2y >= 10, the objective cannot fall below 20, and it equals 20 all along that boundary line. Answer: Minimum Z = 20, attained at every point on the segment joining (10,0) and (0,5).

Question 7

Maximise Z = -2x + 3y subject to x >= 2, x + y >= 6, y >= 0. Does a maximum exist?

Accepted Answer

The region is unbounded upwards, and the point (2, 500) is feasible while giving Z = 1496, which beats every corner value. Answer: No maximum exists; Z increases without limit.

Question 8

Show that the constraints 2x + y <= 4, x + 2y >= 10, x >= 0, y >= 0 admit no feasible solution.

Accepted Answer

The first constraint caps 2x + y at 4 near the origin while the second demands x + 2y be at least 10, and the two shaded half planes never overlap in the first quadrant. Answer: The feasible region is empty, so the problem has no solution.

Practice — Linear Programming

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