Refer to Exercise 30. Minimum value of F is (A) 0 (B) -16 (C) 12 (D) does not exist

NCERT Exemplar · Q42

F

Kerala DhseMCQ· 1mImportance★★★★★

85% · 57/67 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

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.

Note

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.

Tip

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:

"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.
"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).
"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. …

Start your 14-day free trial to unlock the full solution →

Evaluating $F = 3x - 4y$ at the corner points of the Exercise-30 feasible region gives $0$ , $12$ and $-16$ . Since $F$ is bounded on this region, the minimum exists and equals the least corner value, $-16$ — option (B).

Set-up

Exercise 30 gives the objective $F = 3x - 4y$ over the shaded feasible region of Fig. 12.10, whose corner points are $(0,0)$ , $(12,6)$ and $(0,4)$ (with $x \ge 0$ , $y \ge 0$ ). We use the graphical corner-point method: evaluate $F$ at every corner.

Corner-point method

Corner point	$F = 3x - 4y$
$(0,0)$	$3(0) - 4(0) = 0$
$(12,6)$	$3(12) - 4(6) = 12$
$(0,4)$	$3(0) - 4(4) = -16$

These three values — $0$ , $12$ and $-16$ — are exactly the numeric options offered.

Does the minimum exist? …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KEAM 2026Set eng-2026-04174 marksMCQ
$C = 7x + 10y$
›Reveal solutionSolution
Evaluate $C$ at the feasible corner points of the region.

Constraints: $x+y\ge 3$ , $x+2y\ge 4$ , $x,y\ge 0$ . The feasible corners are where the boundary lines meet the axes or each other:

$(4,0)$ : satisfies both; $C=7(4)+10(0)=28$ .

$(0,3)$ : satisfies both; $C=7(0)+10(3)=30$ .

Intersection of $x+y=3$ and $x+2y=4$ : $y=1,\ x=2$ , i.e. $(2,1)$ ; $C=7(2)+10(1)=24$ . …
Start your 14-day free trial to see the full solution →
KEAM 2026Set eng-2026-04184 marksMCQ
$x + y \le 6,\ 2x + y \le 8,\ x \ge 0,\ y \ge 0$
›Reveal solutionSolution

The intersection of the two boundary lines $x+y=6$ and $2x+y=8$ is $(2,4)$ , which satisfies all constraints and is a vertex; the other listed points are infeasible.

Solve $x+y=6$ and $2x+y=8$ : subtracting gives $x=2$ , then $y=4$ , so $(2,4)$ . Check: $2+4=6\le6$ and $2(2)+4=8\le8$ , with $x,y\ge0$ — feasible, and being the meeting point of two active constraints it …
Start your 14-day free trial to see the full solution →
KEAM 2026Set eng-2026-04224 marksMCQ
$Z = x + 2y$
›Reveal solutionSolution

Evaluate $Z$ at the corner points of the feasible region bounded by $2x+3y\le12$ , $x,y\ge0$ .

Corner points: $(0,0)$ , $(6,0)$ and $(0,4)$ . …
Start your 14-day free trial to see the full solution →
KEAM 2025Set eng-2025-04234 marksMCQ
$Q. In the graphical method of a linear programming problem, the optimal solution lies (A) at the centre of the feasible region (B) at a corner point of the feasible region (C) [AMBIGUOUS] (D) [AMBIGUOUS] (E) [AMBIGUOUS]$
›Reveal solutionSolution
[!TLDR]

By the corner-point (fundamental extreme-point) theorem of linear programming, the optimum of the linear objective occurs at a corner point of the feasible region.

Concept

In the graphical method, the constraints define a convex feasible region (a polygon or unbounded polygonal region). A linear objective $Z = ax + by$ has no interior critical point, so its extreme values over a convex polygon are reached at the vertices — this is the Corner-Point Theorem in the NCERT/CBSE linear-programming chapter that KEAM's syllabus follows.

Solution …
Start your 14-day free trial to see the full solution →
KEAM 2025Set eng-2025-04264 marksMCQ
$z = 2x + 3y$
›Reveal solutionSolution
The optimum of a linear objective lies at a corner point; test all four.

Evaluate $z=2x+3y$ :

$(0,0)$ : $0$

$(5,0)$ : $10$

$(4,1)$ : $8+3=11$

$(0,2)$ : $6$ …
Start your 14-day free trial to see the full solution →
KEAM 2024Set eng-2024-06094 marksMCQ
$z = 11x + 7y$
›Reveal solutionSolution

Both $x$ and $y$ have positive coefficients and independent upper bounds, so the optimum is the top corner $(3,2)$ , giving $z=47$ .

The feasible region is the rectangle $0\le x\le3$ , $0\le y\le2$ . Evaluate $z=11x+7y$ at the corners: …
Start your 14-day free trial to see the full solution →
KEAM 2022Set eng-2022-P2-B14 marksMCQ
$z = 50x + 15y$
›Reveal solutionSolution
The objective z = 50x + 15y is largest (1250) at the vertex B(10, 50).

Concept and Intuition

For a linear programming problem the optimum of a linear objective over a convex feasible polygon occurs at a corner (vertex). So we just evaluate z at each vertex and pick the biggest.

Step-by-Step Solution

O(0,0): z = 0.

C(0,60): z = 50(0) + 15(60) = 900.

A(20,0): z = 50(20) + 15(0) = 1000. …
Start your 14-day free trial to see the full solution →

←Previous questionQ41 — The feasible region of a linear programming problem is an unbounded re…→Next questionQ43 — Corner points of the feasible region for an LPP are (0, 2), (3, 0), (6…

Q.Refer to Exercise 30. Minimum value of $F$ is
(A) $0$
(B) $-16$
(C) $12$
(D) does not exist

Concept understanding — Corner Point Theorem

The Corner Point Theorem: Why the Best Answer Hides at the Edges

The Intuition: Why Corners Win

The Precise Statement

Why This Matters for Exams

Set-up

Corner-point method

Does the minimum exist? …

Method: The unbounded-region check before naming an optimum

Steps

Common Mistakes

Concept

Solution …

Concept and Intuition

Step-by-Step Solution

Q.Refer to Exercise 30. Minimum value of FFF is (A) 000 (B) −16-16−16 (C) 121212 (D) does not exist

Set-up

Corner-point method

Does the minimum exist? …

🎓Unlock everything free for 14 days

Q.Refer to Exercise 30. Minimum value of $F$ is
(A) $0$
(B) $-16$
(C) $12$
(D) does not exist

Unlock everything free for 14 days