CLASS 12 · MATHEMATICS PART II · CHAPTER 12
Solve the complete graphical linear-programming exercise by identifying feasible regions, corner points and optimal values.
10 worked answers · short steps first, explanations when you need them
Textbook pages 403–404
Question 1
The feasible region is the triangle in the first quadrant with corner points (0,0), (4,0) and (0,4).
Evaluate Z: at (0,0), Z=0; at (4,0), Z=12; at (0,4), Z=16.
The maximum is Z=16 at (x,y)=(0,4).
Question 2
The corner points are (0,0), (4,0), (2,3), and (0,4). The intersection comes from subtracting x+2y=8 from 3x+2y=12, giving x=2 and y=3.
Evaluate Z at those points: 0, −12, 6 and 16, respectively.
The minimum is Z=−12 at (4,0).
Question 3
The axis vertices are (0,0), (2,0) and (0,3).
Solve 3x+5y=15 and 5x+2y=10: x=20/19 and y=45/19.
The objective values at the four vertices are 0, 10, 9 and 235/19.
The maximum is Z=235/19 at (20/19,45/19).
Question 4
The feasible-region corner points nearest the origin are (3,0), (3/2,1/2), and (0,2). The middle point is where x+3y=3 and x+y=2 intersect.
Their objective values are 9, 7 and 10. The feasible region continues outward, but both coefficients in Z are positive, so moving farther into the region cannot lower these values.
The minimum is Z=7 at (x,y)=(3/2,1/2).
Question 5
Because x,y≥0, 3x+5y is at least 3(x+y).
The first constraint would therefore force 3x+5y≥24, but the second requires 3x+5y≤15. These conditions cannot both hold.
There is no feasible region, so the problem has no feasible solution and no minimum.
Question 6
The second constraint says Z=x+2y≥6.
On the boundary x+2y=6, the segment from (0,3) to (6,0) is in the first quadrant. Along it, 2x+y ranges from 3 to 12, so the first constraint is also satisfied.
The minimum is 6 at every point on the segment x+2y=6 from (0,3) to (6,0); therefore there are infinitely many minimum points.
Question 7
The corner points are (60,0), (120,0), (60,30), and (40,20). They come from the axes and intersections of the boundary lines.
Since Z=5(x+2y), the constraint x+2y≤120 gives Z≤600. At (60,0), Z=300; at (40,20), Z=400; at (60,30) and (120,0), Z=600.
The minimum is 300 at (60,0). The maximum is 600 at every feasible point on x+2y=120 between (60,30) and (120,0).
Question 8
Rewrite 2x−y≤0 as y≥2x. The corner points are (0,50), (0,200), (50,100), and (20,40).
The objective values there are 100, 400, 250, and 100.
On the boundary x+2y=100, the feasible segment from (0,50) to (20,40) satisfies all constraints, so every point on it gives 100.
The minimum is 100 at every point on the segment from (0,50) to (20,40). The maximum is 400 at (0,200).
Question 9
Fix x=3. Any y≥3 satisfies x+y≥5 and x+2y≥6.
For these feasible points, Z=−3+2y, which increases without limit as y increases.
There is no maximum value; Z is unbounded above on the feasible region.
Question 10
The first inequality gives y≥x+1.
The second gives y≤x. No number y can be both at least x+1 and at most x.
The constraints have no common feasible point, so there is no feasible solution and no maximum.