Q.For the linear programming problem (LPP), the objective function is Z=4x+3y and the feasible region determined by a set of constraints is shown in the graph: (Note: The figure is not to scale.) Which of the following statements is true?
(A) Maximum value of Z is at R(40,0).
(B) Maximum value of Z is at Q(30,20).
(C) Value of Z at R(40,0) is less than the value at P(0,40).
(D) The value of Z at Q(30,20) is less than the value at R(40,0).
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Corner Point Theorem
The Corner Point Theorem: Why the Best Answer Hides at the Edges
Imagine maximising profit for a factory that makes two products, with limited raw materials, machine hours, and labour. Every combination that doesn't break a limit is a feasible solution. Plot them all on a graph and they form a shape — always a polygon if your constraints are straight lines.
Where is the best (maximum profit) point? You might think anywhere inside the shape. But the Corner Point Theorem says something surprising: the best point is always at a corner — a vertex of the polygon. Never floating in the middle of an edge or inside.
This theorem is the backbone of linear programming — the method for solving optimisation problems with straight-line constraints.
The Intuition: Why Corners Win
Think of profit as a line you slide across the polygon; each position represents a profit level. You push the line as far as possible (higher profit) while still touching the polygon. The last point of contact before the line escapes is always a corner.
Why? Because both the profit line and the polygon's edges are straight, and the farthest point in any straight-line direction from a polygon is always a vertex. This holds for any flat-sided shape.
Solving a linear programming problem by hand, you only need to check the corners — usually just 3–5 points, not the infinite points inside.
The Precise Statement
Corner Point Theorem (Fundamental Theorem of Linear Programming):
If a linear programming problem has an optimal solution, then that optimal solution occurs at at least one corner point (vertex) of the feasible region.
Three key parts:
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"If it has an optimal solution" — sometimes the problem is unbounded (profit increases forever) or infeasible (no point satisfies all constraints). The theorem applies only when a best answer exists.
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"At least one corner point" — several corners can give the same optimal value. If the profit line is parallel to an edge, every point on that edge is optimal, including both endpoints (corners).
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"Of the feasible region" — the polygon formed by all constraints. Corners are where two constraint lines intersect.
Why This Matters for Exams
To solve a linear programming problem:
- Find all corner points (solve pairs of constraint equations).
- Plug each corner into the objective function.
- Pick the best value.
The theorem guarantees you haven't missed a better answer hiding in the middle. …
Concept: Corner Point Theorem – For a linear programming problem with a bounded feasible region, the optimal value of the objective function occurs at one of the corner points of the region.
Step 1: Identify the corner points from the graph.
The feasible region has vertices P(0,40), Q(30,20), and R(40,0).
Step 2: Evaluate Z=4x+3y at each corner point.
- At P(0,40): Z=4(0)+3(40)=120
- At Q(30,20): Z=4(30)+3(20)=120+60=180
- At R(40,0): Z=4(40)+3(0)=160
Step 3: Compare the values.
Z is maximum at Q(30,20) with value 180. …
The Corner Point Theorem says the optimum of a linear objective over a convex polygon occurs at a vertex. Evaluating Z=4x+3y at the three corner points gives Z(P)=120, Z(Q)=180, Z(R)=160, so the maximum is at Q(30,20) and option (B) is correct.
The Corner Point Theorem (also called the Fundamental Theorem of Linear Programming) is the key idea here. It states that if a linear programming problem has an optimal solution and the feasible region is a bounded convex polygon, then the optimum occurs at one of the vertices (corner points) of that region. This saves us from checking every single point inside the region — we only need to evaluate the objective function at the corners.
In the given graph, the feasible region is a triangle with vertices P(0,40), Q(30,20), and R(40,0). Let’s evaluate Z=4x+3y at each.
-
At P(0,40):
Z=4(0)+3(40)=0+120=120
-
At Q(30,20):
Z=4(30)+3(20)=120+60=180
-
At R(40,0):
Z=4(40)+3(0)=160+0=160
Now compare the values: 120, 180, 160. The largest is 180 at Q(30,20). So the maximum value of Z is at Q. …
Method: Corner-Point Test for the Optimal Vertex
Use this when a bounded feasible region's vertices are known (or readable from a graph) and you must decide which corner optimises Z=ax+by, or judge which comparison statement about the corners is true.
Steps
Step 1: List every corner point.
By the Corner-Point Theorem, the optimum of a linear objective on a bounded region always sits at a vertex — so the interior can be ignored. Collect all vertices of the region.
Step 2: Evaluate Z=ax+by at each corner.
Substitute each vertex's coordinates into Z and tabulate the values. Do this for every corner — never assume the answer from a coordinate.
Step 3: Compare, then test the claim. …
Common Mistakes
Mistake 1: Assuming the corner with the largest x gives the maximum.
Why it's wrong: R(40,0) has the biggest x, but Z(R)=160 while Z(Q)=4(30)+3(20)=180 — the interior-edge corner Q wins because y carries weight 3. Correct approach: evaluate Z=4x+3y at every corner and compare, don't judge by a single coordinate.
Mistake 2: Mis-comparing two corner values in the statements. …
Showing the 12 most recent of 14 on this concept.
- GUJCET 2025Set 031 markMCQQ.The coordinates of the corner points of the bounded feasible region are (0,0), (0,40), (20,40), (60,20), (60,0). The maximum of the objective function z=40x+30y is _____. (A) 2000 (B) 3400 (C) 2400 (D) 3000
›Reveal solutionSolution
For a bounded LP region the optimum sits at a corner; test all corners.
With z=40x+30y:
- (0,0)→0
- (0,40)→1200
- (20,40)→800+1200=2000 …
- GUJCET 2023Set 091 markMCQQ.The coordinates of the corner points of the bounded feasible region are (0,10), (5,5), (15,15) and (0,20). The maximum of the objective function Z=10x+20y is : (A) 600 (B) 550 (C) 400 (D) 450
›Reveal solutionSolution
[!TLDR]
[L]=M1L2T−2A−2.
Concept
Self inductance relates to stored magnetic energy U=21LI2 (NCERT/GSEB electromagnetic induction), so L=I22U.
Solution
Energy has dimensions [ML2T−2] and current [A]: …
- GUJCET 2019Set 171 markMCQQ.The coordinates of the corner points of the bounded feasible region are (10,0), (2,4), (1,5) and (0,8). The maximum of objective function z=60x+10y is . (A) 800 (B) 600 (C) 700 (D) 110
›Reveal solutionSolution
Checking corner points, z=60x+10y is largest at (10,0) giving 600.
Concept: The maximum of a linear objective on a bounded feasible region occurs at a corner.
- (10,0):600 …
- GUJCET 2022Set 081 markMCQQ.Corner points of the feasible region of objective function Z=3x+9y of a linear programming problem are (0,10), (5,5), (15,15) and (0,20). Minimum value of Z is ______. (A) 70 (B) 90 (C) 50 (D) 60
›Reveal solutionSolution
In LP the optimum of a linear objective occurs at a corner point — just test them all.
Concept. Evaluate Z=3x+9y at every corner and pick the smallest.
Solution.
- (0,10):0+90=90
- (5,5):15+45=60 …
- GUJCET 2026Set x1 markMCQQ.The coordinates of the corner points of the bounded feasible region are (0,10),(5,5),(15,15),(0,20). The minimum of the objective function z=3x+9y is ______ (A) 180 (B) 30 (C) 90 (D) 60
›Reveal solutionSolution
[!TLDR] The minimum of z=3x+9y over the corner points is 60 at (5,5).
Concept
In linear programming, a linear objective attains its optimum over a bounded feasible region at a corner (vertex). So evaluate z at every corner point and choose the smallest for the minimum.
Solution …
- GUJCET 2024Set 131 markMCQQ.The coordinates of the corner points of the bounded feasible region are (0,6),(3,3),(9,9),(0,12). The maximum of the objective function z=6x+12y is : (A) 166 (B) 152 (C) 144 (D) 162
›Reveal solutionSolution
Evaluate z=6x+12y at every corner point; the maximum is the answer.
Steps.
- (0,6): 0+72=72
- (3,3): 18+36=54
- (9,9): 54+108=162 …
- GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQQ.For a linear programming problem, the objective function Z=510x+675y has the corner points of the bounded feasible region (0,0), (300,0), (180,120) and (0,240). The maximum value of Z is ____.(a) 1,72,800(b) 1,62,000(c) 1,53,000(d) 1,70,000
›Reveal solutionSolution
Evaluate Z at each corner point and pick the largest.
Z=510x+675y: (0,0)→0; (300,0)→153000; (180,120)→91800+81000=172800; (0,240)→162000.
…
- GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQQ.In a linear programming problem, Z=px+qy, p,q>0. At the corner points (0,10) and (5,5) of the bounded feasible region, the values of Z are 90 and 60 respectively. Then the relation between p and q = ____.(a) p=3q(b) q=2p(c) q=3p(d) p=2q
›Reveal solutionSolution
Plug both corner points into Z=px+qy to get two equations in p,q, then solve.
At (0,10): 10q=90⇒q=9.
…
- GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQQ.The corner points of the feasible region of the objective function Z=−50x+20y are (0,5), (0,3), (1,0) and (6,0). Then the minimum value of Z = ____.(a) −100(b) −300(c) −200(d) −500
›Reveal solutionSolution
Evaluate Z at every corner point of the bounded feasible region; the minimum is the smallest value.
Z(0,5)=0+100=100
Z(0,3)=0+60=60
Z(1,0)=−50+0=−50
Z(6,0)=−300+0=−300
…
- GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQQ.For linear programming problem, the objective function is Z=3x+2y. If the corner points of the bounded feasible region are (12,0),(4,2),(1,5) and (0,10), then the maximum value of Z is ______.(a) 36(b) 46(c) 13(d) 56
›Reveal solutionSolution
In an LP problem the optimum of a linear objective function always occurs at a corner point of the feasible region.
Evaluate Z=3x+2y at each corner point:
(12,0):Z=36
(4,2):Z=12+4=16
(1,5):Z=3+10=13 …
- GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQQ.The corner points of the feasible region determined by the following system of linear inequalities 2x+y≤10, x+3y≤15, x,y≥0 are (0,0), (5,0), (3,4) and (0,5). Let Z=qx+py where p,q>0, condition on p and q so that the maximum of Z occurs at both (3,4) and (0,5) is ___.(a) q=p(b) q=2p(c) q=3p(d) p=3q
›Reveal solutionSolution
If the maximum of Z occurs at two adjacent corner points, Z is equal there.
Z=qx+py. Set Z(3,4)=Z(0,5):
…
- GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQQ.The corner points of feasible region determined by the system of linear constraints are (2,72), (15,20), (40,15). Let Z=6x+3y be the objective function. Minimum of Z occurs at ___.(a) (2,72)(b) (15,20)(c) (40,15)(d) (0,0)
›Reveal solutionSolution
Evaluate the objective at each corner point and pick the smallest.
Z(2,72)=12+216=228.
Z(15,20)=90+60=150.
Z(40,15)=240+45=285.
…
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