Where Mindful Learning Begins
Where Mindful Learning Begins
This Class 12 Maths NCERT Solutions Chapter 12 Ex 12.1 page covers all 10 questions, solved step-by-step, using the graphical Corner Point Method. Exercise 12.1 is the only exercise in this chapter — there's no 12.2 or Miscellaneous Exercise — so every idea CBSE expects you to know about Linear Programming shows up somewhere in these 10 questions.
The method is always the same three steps: graph the constraints to find the feasible region, locate its corner points by solving pairs of boundary equations simultaneously, then evaluate the objective function Z at each corner point. Questions 1, 3 and 5 are the standard bounded-region case where the largest or smallest value among the corner points is simply the answer. Question 4 introduces an unbounded feasible region, where you must additionally check that no better value exists further out before declaring an optimum. Questions 6, 7 and 8 deal with the "multiple optimal solutions" case — where the objective function happens to run parallel to one edge of the feasible region, so the optimum is achieved along an entire line segment, not just a single point; this is a favourite CBSE "show that" question. Questions 9 and 10 close the exercise with two edge cases every student needs to recognise on sight: a region where the objective function has no maximum at all, and a problem with no feasible region whatsoever.
The feasible region is the triangle bounded by x+y=4 and the two axes, with corner points O(0,0), A(4,0) and B(0,4). This region is bounded, so we evaluate Z at each corner.
| Corner point | Z = 3x + 4y |
|---|---|
| O (0, 0) | 0 |
| A (4, 0) | 12 |
| B (0, 4) | 16 ← Maximum |
Solving x+2y=8 and 3x+2y=12 simultaneously (subtracting) gives 2x=4 \Rightarrow x=2,\ y=3. The feasible region is bounded, with corner points O(0,0), A(4,0), B(2,3), C(0,4).
| Corner point | Z = −3x + 4y |
|---|---|
| O (0, 0) | 0 |
| A (4, 0) | −12 ← Minimum |
| B (2, 3) | 6 |
| C (0, 4) | 16 |
Solving 3x+5y=15 and 5x+2y=10 simultaneously gives x=\dfrac{20}{19},\ y=\dfrac{45}{19}. Corner points: O(0,0), A(2,0), B(20/19, 45/19), C(0,3).
| Corner point | Z = 5x + 3y |
|---|---|
| O (0, 0) | 0 |
| A (2, 0) | 10 |
| B (20/19, 45/19) | 235/19 ← Maximum |
| C (0, 3) | 9 |
Solving x+3y=3 and x+y=2 simultaneously gives y=\dfrac12,\ x=\dfrac32. Corner points on the boundary: A(3,0), B(3/2, 1/2), C(0,2), and the feasible region is unbounded above.
| Corner point | Z = 3x + 5y |
|---|---|
| A (3, 0) | 9 |
| B (3/2, 1/2) | 7 ← smallest |
| C (0, 2) | 10 |
Since the region is unbounded, we check whether the open half-plane 3x+5y<7 shares any point with the feasible region. It does not, so 7 is confirmed as the true minimum.
Solving x+2y=10 and 3x+y=15 simultaneously gives x=4,\ y=3. Corner points: O(0,0), A(5,0), B(4,3), C(0,5).
| Corner point | Z = 3x + 2y |
|---|---|
| O (0, 0) | 0 |
| A (5, 0) | 15 |
| B (4, 3) | 18 ← Maximum |
| C (0, 5) | 10 |
The two boundary lines 2x+y=3 and x+2y=6 intersect exactly at (0,3), and for x>0 the line x+2y=6 lies above 2x+y=3.
So the binding constraint throughout the feasible region is x+2y\ge6, whose boundary runs from (0,3) to (6,0).
| Corner point | Z = x + 2y |
|---|---|
| (0, 3) | 6 ← Minimum |
| (6, 0) | 6 ← Minimum |
Notice that the objective function Z=x+2y is exactly the left-hand side of the constraint x+2y\ge6. So for every point on the boundary segment joining (0,3) and (6,0) — where x+2y=6 holds — Z equals 6.
Solving the boundary lines pairwise:
x+2y=120 with x=2y gives (60, 30).
x+y=60 with x=2y gives (40, 20).
x+y=60 with y=0 gives (60, 0).
x+2y=120 with y=0 gives (120, 0).
The feasible region is the bounded quadrilateral with these four corners.
| Corner point | Z = 5x + 10y |
|---|---|
| (60, 0) | 300 ← Minimum |
| (120, 0) | 600 ← Maximum |
| (60, 30) | 600 ← Maximum |
| (40, 20) | 400 |
Since Z=5x+10y=5(x+2y), and both (120,0) and (60,30) lie on the constraint line x+2y=120, Z equals 600 along the entire edge joining them.
Solving the boundary lines pairwise:
x+2y=100 with y=2x gives (20, 40).
y=2x with 2x+y=200 gives (50, 100).
x+2y=100 with x=0 gives (0, 50).
2x+y=200 with x=0 gives (0, 200).
These four points form the bounded feasible quadrilateral.
| Corner point | Z = x + 2y |
|---|---|
| (0, 50) | 100 ← Minimum |
| (20, 40) | 100 ← Minimum |
| (50, 100) | 250 |
| (0, 200) | 400 ← Maximum |
Both (0,50) and (20,40) lie on the constraint line x+2y=100, so — exactly like the objective function itself — Z stays at 100 along the entire edge between them.
Solving x=3 with x+y=5 gives (3, 2); solving x+y=5 with x+2y=6 gives (4, 1). The feasible region is unbounded, extending upward along the line x=3.
| Corner point | Z = −x + 2y |
|---|---|
| (3, 2) | 1 |
| (4, 1) | −2 |
Checking the open half-plane -x+2y>1: the point (3, 3) is feasible (it satisfies all three constraints) and gives Z=-3+6=3>1. So the feasible region has points with Z arbitrarily larger than any corner-point value.
The constraint x-y\le-1 rearranges to y\ge x+1. The constraint -x+y\le0 rearranges to y\le x.
These two conditions require x+1\le y\le x simultaneously — but x+1 is always strictly greater than x, so no value of y can satisfy both at once.
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