Consider the following Linear Programming Problem: Minimise $Z = x + 2y$ Subject to $2x + y \geq 3$, $x + 2y \geq 6$, $x, y \geq 0$. Show graphically that the minimum of $Z$ occurs at more than two points.

Consider the following Linear Programming Problem: Minimise Z = x + 2y Subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y

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Z = x + 2y

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The minimum value is $Z = 6$ , attained at every point of the line segment joining $(0,3)$ and $(6,0)$ — infinitely many points, hence at more than two.

Draw the constraint boundaries.

$2x + y = 3$ meets the axes at $(1.5, 0)$ and $(0, 3)$ .
$x + 2y = 6$ meets the axes at $(6, 0)$ and $(0, 3)$ .

Both inequalities are " $\ge$ ", so the feasible region lies above/right of each line, with $x, y \ge 0$ . The region is unbounded, and its two corner points are $(0,3)$ and $(6,0)$ .

Evaluate $Z = x + 2y$ at the corners.

Z(0,3) = 0 + 2(3) = 6, \qquad Z(6,0) = 6 + 2(0) = 6.

Both corners give the same value, $Z = 6$ .

Why the minimum repeats along a whole edge. …

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

KCET 2026Set UNKNOWN1 markMCQ
$(0, 10)$
›Reveal solutionSolution

If $z = px+qy$ has its maximum simultaneously at $(15,15)$ and $(0,20)$ , the value of $z$ at both points must be equal — set the two expressions equal and solve for the ratio $q:p$ .

Step 1 — Set up the equal-value condition

Since the maximum of $z$ occurs at both $(15,15)$ and $(0,20)$ , the objective function must give the same value there:
$z(15,15) = z(0,20).$
Step 2 — Substitute the coordinates
$p(15) + q(15) = p(0) + q(20)$ $15p + 15q = 20q.$
Step 3 — Solve for the relation between $p$ and $q$
$15p = 20q - 15q = 5q$
$15p = 5q \implies q = 3p.$ …
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KCET 2026Set UNKNOWN1 markMCQ
$Z = ax + by$
›Reveal solutionSolution

A linear function is constant along any line segment joining two points at which it takes equal values, so if two corner points share the maximum, the whole edge between them does too — giving infinitely many optimal points.

Step 1 — Recall the theorem on optimal solutions in LPP

For a linear objective function $Z = ax+by$ over a convex feasible region, if $Z$ attains the same maximum value $M$ at two distinct corner points $P_1$ and $P_2$ , then $Z$ attains the value $M$ at every point on the line segment $P_1P_2$ .

Step 2 — Why this is true

Any point on the segment joining $P_1=(x_1,y_1)$ and $P_2=(x_2,y_2)$ can be written as
$(x,y) = \big(\lambda x_1 + (1-\lambda)x_2,\ \lambda y_1 + (1-\lambda)y_2\big), \quad 0 \le \lambda \le 1.$
Substituting into $Z=ax+by$ and using $ax_1+by_1 = ax_2+by_2 = M$ gives …
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COMEDK 2025Set 2025-A1 markMCQ
$(0,3),(1,1)$
›Reveal solutionSolution
If a linear objective attains its minimum at two corner points, it is constant along the edge joining them; equating $Z$ at $(3,0)$ and $(1,1)$ gives $2p=q$ , i.e. $p=\dfrac{q}{2}$ — option (C).

Concept

In a linear programming problem the feasible region is a convex polygon and $Z=px+qy$ is linear, so its optimum occurs at a corner point. If the same minimum is reached at two different corners, then every point on the segment between them gives that identical minimum value — i.e. $Z$ is constant along that edge. That single fact provides one equation relating $p$ and $q$ .

Solution

The minimum is attained at $(3,0)$ and $(1,1)$ , so evaluate $Z$ at each and set them equal.

At $(3,0)$ :

$Z=p(3)+q(0)=3p.$

At $(1,1)$ :

$Z=p(1)+q(1)=p+q.$

Both are the minimum, so …
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KCET 2022Set C-41 markMCQ
$z = 4x + 6y$
›Reveal solutionSolution

The minimum value of $z$ is achieved at two adjacent corner points, and by the corner-point theorem it is then achieved at every point of the line segment joining them — infinitely many points.

Step 1 — Evaluate $z=4x+6y$ at each corner point.

Corner $z=4x+6y$
$(0,2)$ $0+12=\mathbf{12}$
$(3,0)$ $12+0=\mathbf{12}$
$(6,0)$ $24$
$(6,8)$ $24+48=72$
$(0,5)$ $30$

Step 2 — Identify the minimum. The smallest value is $z_{\min}=12$ , and it occurs at two corner points: $(0,2)$ and $(3,0)$ . …
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Corner	$z=4x+6y$
$(0,2)$	$0+12=\mathbf{12}$
$(3,0)$	$12+0=\mathbf{12}$
$(6,0)$	$24$
$(6,8)$	$24+48=72$
$(0,5)$	$30$

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Q.Consider the following Linear Programming Problem: Minimise $Z = x + 2y$ Subject to $2x + y \geq 3$ , $x + 2y \geq 6$ , $x, y \geq 0$ . Show graphically that the minimum of $Z$ occurs at more than two points.

Concept understanding — Multiple Optimal Solutions

Multiple Optimal Solutions

A concrete example

Why it happens

How to recognise it

Why it matters …

Method: Detecting Infinitely Many Optima (Objective Parallel to an Edge)

Steps

Common Mistakes

Step 1 — Set up the equal-value condition

Step 2 — Substitute the coordinates

Step 3 — Solve for the relation between $p$ and $q$

Step 1 — Recall the theorem on optimal solutions in LPP

Step 2 — Why this is true

Concept

Solution

Q.Consider the following Linear Programming Problem: Minimise Z=x+2yZ = x + 2yZ=x+2y Subject to 2x+y≥32x + y \geq 32x+y≥3, x+2y≥6x + 2y \geq 6x+2y≥6, x,y≥0x, y \geq 0x,y≥0. Show graphically that the minimum of ZZZ occurs at more than two points.

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Q.Consider the following Linear Programming Problem: Minimise $Z = x + 2y$ Subject to $2x + y \geq 3$ , $x + 2y \geq 6$ , $x, y \geq 0$ . Show graphically that the minimum of $Z$ occurs at more than two points.

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