Q.Determine the maximum value of $Z = 11x + 7y$ subject to the constraints: $2x + y \le 6$, $x \le 2$, $x \ge 0$, $y \ge 0$.

Step 1 — Read the feasible region from the figure.

The shaded region is bounded by the $Y$-axis on the left, by $x+y=5$ above and by $x+3y=9$ below, i.e.

Its corner points are the three points printed in the figure: $(0,5)$, $(0,3)$ and the intersection of the two lines.

Step 2 — Confirm the intersection point.

Solve $x+y=5$ and $x+3y=9$ simultaneously. Subtracting the first from the second:

So the lines meet at $(3,2)$, exactly as the figure states.

Step 3 — Corner-point theorem.

For a linear objective function over a closed, bounded convex feasible region, the maximum (and minimum) is attained at a vertex of the region. So we need only evaluate $Z$ at the three corners.

Step 4 — Evaluate $Z=11x+7y$.

Step 5 — Compare.

$47 > 35 > 21$, so the maximum value of $Z$ is $47$, occurring at $(3,2)$.

Why not (C) $(5,0)$? It is merely the x-intercept of $x+y=5$; it fails the constraint $x+3y\ge 9$ (since $5+0=5<9$) and, as the figure shows, is not in the shaded region. It is a trap option: $Z(5,0)=55$ would be larger, but the point is infeasible.

ANSWER: D

Step 1 — Write down the LPP.

Maximise $z = 3x + 4y$ subject to

Step 2 — Why corner points suffice.

The Corner Point Theorem states: if the feasible region of an LPP is bounded (a convex polygon), then the objective function attains both its maximum and its minimum at a vertex (corner point) of that region. So we never need to check the interior — only the corners.

Step 3 — Find every corner of the feasible region.

The boundary lines are $x + y = 40$, $x + 2y = 60$, $x = 0$ and $y = 0$.

• Origin: $(0, 0)$ — satisfies both constraints. ✓
• On the $x$-axis ($y = 0$): $x + y \le 40 \Rightarrow x \le 40$; $x + 2y \le 60 \Rightarrow x \le 60$. The binding one is $x \le 40$, giving the corner $(40, 0)$. ✓
• On the $y$-axis ($x = 0$): $y \le 40$ and $2y \le 60 \Rightarrow y \le 30$. The binding one is $y \le 30$, giving the corner $(0, 30)$. ✓
• Intersection of the two lines: solve

Subtracting (i) from (ii): $y = 20$. Then from (i): $x = 40 - 20 = 20$. Corner $(20, 20)$. ✓ (Check: $20+20=40 \le 40$ ✓ and $20+40=60 \le 60$ ✓, and both coordinates are $\ge 0$, so it is genuinely feasible.)

Step 4 — Evaluate $z = 3x + 4y$ at each corner.

Step 5 — Read off the maximum.

The largest value is $\mathbf{140}$, attained at $(20, 20)$. Note the trap: both "easy" axis corners give $120$ (option B) — the true optimum lies at the intersection of the two constraints, which is exactly where a corner-point check earns its keep.

ANSWER: C

Maximise $Z=3x+4y$ subject to

Find the corner points of the feasible region.

• $(0,0)$: $Z=0$.
• $x$-axis ($y=0$): the binding limit is $2x\le 104\Rightarrow x=52$, giving $(52,0)$, $Z=156$.
• $y$-axis ($x=0$): the binding limit is $2y\le 76\Rightarrow y=38$, giving $(0,38)$, $Z=152$.
• Intersection of $x+2y=76$ and $2x+y=104$: multiply the first by $2$ to get $2x+4y=152$, then subtract the second: $3y=48\Rightarrow y=16$, and $x=76-2(16)=44$. Point $(44,16)$.

At $(44,16)$:

Comparing all corners, the maximum value is $196$.

Step 1 — Write down the feasible region.

Maximise $z = x + 4y$ subject to

Step 2 — Test the two constraints together.

Put $t = x + 2y$. The constraints say simultaneously

which is impossible for any real $t$. The half-plane $x+2y\le 2$ lies below the line $x+2y=2$ and the half-plane $x+2y\ge 8$ lies above the parallel line $x+2y=8$; the two lines are parallel (same normal vector $(1,2)$) and never meet, so the half-planes are disjoint.

Step 3 — Conclusion for the LPP.

The feasible region $S = \varnothing$. A linear programming problem with an empty feasible region is called infeasible: there are no corner points, so the corner-point method cannot be applied and no maximum value of $z$ exists.

Hence $z_{\max}$ is not $4$, $8$ or $16$ — the objective function is never evaluated at all.

ANSWER: D

1. Name the boundary lines

• The line of negative slope through $(0,7)$ and $(7,0)$ is $\dfrac{x}{7}+\dfrac{y}{7}=1$, i.e. $\;x+y=7$.
• The line of positive slope through $(-3,0)$ and $(0,2)$ is $\dfrac{x}{-3}+\dfrac{y}{2}=1 \Rightarrow -2x+3y=6$, i.e. $\;2x-3y+6=0$.

Check their intersection: solving $x+y=7$ with $2x-3y+6=0$ gives $2x-3(7-x)+6=0 \Rightarrow 5x=15 \Rightarrow x=3,\ y=4$ — the point $B(3,4)$ marked on the figure. ✓ (Good — the transcription is consistent.)

2. The region is in the first quadrant

The shading lies entirely to the right of the $y$-axis and above the $x$-axis, so $x \ge 0$ and $y \ge 0$. Every option carries these, so they do not discriminate.

3. Test an interior point — the origin

The shaded quadrilateral $O(0,0), C(0,2), B(3,4), A(7,0)$ has the origin as a vertex, so pick a point clearly inside, say $(1,1)$ (and $O$ itself for the boundary check):

• Against $x+y=7$: at $(1,1)$, $x+y = 2 \le 7$. The region is on the origin side of the line.

• Against $2x-3y+6=0$: at $(1,1)$, $2(1)-3(1)+6 = 5 > 0$. The region is again on the origin side.

4. Sanity-check a vertex on each line

$A(7,0)$: $x+y=7$ (on the boundary ✓) and $2(7)-0+6 = 20 \ge 0$ ✓.

$C(0,2)$: $2(0)-3(2)+6 = 0$ (on the boundary ✓) and $0+2 \le 7$ ✓.

Both vertices satisfy the system, confirming the constraint set.

ANSWER: D

[!TLDR]

The feasible region is the triangle with corners (0,2), (2,0), (3,0), matching option (B).

Concept

Each linear inequality defines a half-plane; the feasible region is their intersection in the first quadrant — core CBSE/NCERT Class 12 'Linear Programming' work.

Solution

Boundary lines:

• $x+y=2$ passes through $(2,0)$ and $(0,2)$.
• $2x+3y=6$ passes through $(3,0)$ and $(0,2)$.

Both lines intersect at $(0,2)$.

Apply the inequalities:

• $x+y\ge 2$: on/above the first line.
• $2x+3y\le 6$: on/below the second line.
• $x\ge 0,\ y\ge 0$: first quadrant.

Along the x-axis ($y=0$): $x\ge 2$ and $x\le 3$, giving the segment from $(2,0)$ to $(3,0)$. Together with the common point $(0,2)$, the region is the triangle with vertices $(0,2),(2,0),(3,0)$.

Check an interior point $(2.5,0.1)$: $x+y=2.6\ge2$ ✓ and $2x+3y=5.3\le6$ ✓.

This is exactly the hatched triangle described in option (B).

[!ANSWER]

(B) Triangle with corners $(0,2),(2,0),(3,0)$

We are given a linear programming problem with objective $Z = 80x + 120y$ and constraints:

The question asks for a corner point $P$ of the feasible region. The feasible region is the set of all points satisfying all constraints. Its corners occur where two boundary lines intersect (including the axes). We need to find which of the given options is actually a corner.

Concept & Intuition:

In linear programming, the feasible region is a polygon (here a quadrilateral). Its vertices are found by solving each pair of boundary equations. The axes $x=0$ and $y=0$ are also boundaries. So we check intersections of:

• $x+3y=30$ with $3x+4y=60$
• Each line with the axes
• The axes themselves

Then we verify which of the given points is among these vertices.

Step-by-step solution:

1. Intersection of the two constraint lines Solve:

Multiply the first equation by 3: $3x + 9y = 90$. Subtract the second:

$(3x+9y) - (3x+4y) = 90 - 60 \Rightarrow 5y = 30 \Rightarrow y = 6$.

Then $x = 30 - 3(6) = 12$. So intersection is $(12,6)$.

But note: This point is not among the options. However, it is a corner of the feasible region.

2. Intersection of $x+3y=30$ with the y-axis ($x=0$)

   $0 + 3y = 30 \Rightarrow y = 10$. Point: $(0,10)$.

   Intersection with the x-axis ($y=0$): $x = 30$. Point: $(30,0)$.

   This gives option (D) $(30,0)$.

3. Intersection of $3x+4y=60$ with the y-axis ($x=0$)

   $0 + 4y = 60 \Rightarrow y = 15$. Point: $(0,15)$.

   This gives option (A) $(0,15)$.

   Intersection with the x-axis ($y=0$): $3x = 60 \Rightarrow x = 20$. Point: $(20,0)$.

   This gives option (B) $(20,0)$.

4. Check option (C) $(6,12)$

   Does it lie on either line?

   For $x+3y=30$: $6 + 36 = 42 \neq 30$.

   For $3x+4y=60$: $18 + 48 = 66 \neq 60$.

   So it is not on either boundary line — but could it be a corner? Corners occur only at intersections of boundaries. Since (6,12) satisfies neither equality, it is inside the feasible region, not a corner.

   Wait — let’s verify feasibility: $6+3(12)=42 > 30$ violates the first constraint. So (6,12) is actually outside the feasible region. Thus it cannot be a corner.

5. Identify all actual corners

   The feasible region is bounded by:

   • Origin $(0,0)$
   • $(20,0)$ from $3x+4y=60$ and x-axis
   • $(12,6)$ from intersection of the two lines
   • $(0,10)$ from $x+3y=30$ and y-axis But note: $(0,15)$ is on the line $3x+4y=60$ and the y-axis, but is it feasible? Check $x+3y \leq 30$: $0+45=45 > 30$, so it violates the first constraint. Hence $(0,15)$ is not in the feasible region. Similarly, $(30,0)$ violates $3x+4y \leq 60$ because $90 > 60$. So the actual corners are $(0,0)$, $(20,0)$, $(12,6)$, $(0,10)$.

6. Match with options

   Among the given choices, only $(20,0)$ is a genuine corner of the feasible region.

   So the correct option is (B).

The key idea is that in linear programming, when the minimum occurs at two distinct corner points of the feasible region, the objective function is constant along the line segment joining them. This happens because the level lines of $z = px + qy$ are parallel to that edge of the feasible region.

Think of it this way: the feasible region is a convex polygon. The minimum of a linear function over a convex polygon occurs at a corner. If two corners give the same minimum value, then every point on the edge between them also gives that same minimum — the objective function’s contour lines are aligned with that edge.

Here, the two points are $(3,0)$ and $(1,1)$. The line joining them has a certain slope. For $z$ to be constant along that line, the gradient vector $(p, q)$ must be perpendicular to the direction vector of the line (since $z$ changes fastest perpendicular to the contour lines). Equivalently, the slope of the contour line $px + qy = \text{constant}$ must match the slope of the edge.

Let’s work it out step by step.

1. Find the direction vector of the edge joining $(3,0)$ and $(1,1)$. The vector from $(3,0)$ to $(1,1)$ is

So the slope of this edge is $\frac{1}{-2} = -\frac{1}{2}$.

2. Write the condition for $z$ to be constant along this edge. For any two points on the edge, $z$ must be equal. In particular, at $(3,0)$ and $(1,1)$:

This gives:

3. Solve the equation.

4. Interpret the result. Since $p, q > 0$, this condition is valid. It means the objective function’s contour lines have slope $-\frac{p}{q} = -\frac{1}{2}$, exactly matching the slope of the edge between the two optimal points. Hence the entire edge is optimal, and the minimum occurs at both corners (and everywhere between them).

1. Corner-point method. For a bounded feasible region, the optimum of a linear objective occurs at a vertex. The four options are precisely the vertices of the region shown.

2. Evaluate $z = 3x + 9y$ at each vertex.

3. Compare. The smallest value is $\boxed{60}$, attained at $(5,5)$. (This is the standard NCERT feasible region $x+3y\le 60,\; x+y\ge 10,\; x\le y,\; x,y\ge 0$, whose vertices are exactly these four points; its minimum of $3x+9y$ is $60$ at $(5,5)$ and its maximum $180$ at both $(15,15)$ and $(0,20)$.)

ANSWER: A

The key idea is that a shaded region in a linear programming problem is the intersection of half-planes. Each boundary line corresponds to an equation; the inequality sign tells us which side of that line is included. To decide the sign, pick a test point on the shaded side and see whether it satisfies the inequality.

Let’s identify the three boundary lines visible in the figure (the axes are also boundaries, since $x\ge0$ and $y\ge0$ are given in every option).

1. Line through $(0,4)$ and $(5,0)$

   Its equation: intercept form $\frac{x}{5} + \frac{y}{4} = 1$, which simplifies to $4x + 5y = 20$.

   The shaded region lies above this line (away from the origin). Test the origin $(0,0)$: $4(0)+5(0)=0$, which is not $\ge 20$, so the origin is not in the shaded region. Hence the inequality is $4x+5y \ge 20$.

2. Line through $(0,3)$ and $(10,0)$

   Equation: $\frac{x}{10} + \frac{y}{3} = 1$, i.e. $3x + 10y = 30$.

   The shaded region lies below this line (toward the origin). Test $(0,0)$: $0 \le 30$, so the origin satisfies $3x+10y \le 30$. That matches the shading.

3. Vertical line $x=6$

   The shaded region is to the left of this line (smaller $x$). Test a point like $(0,0)$: $0 \le 6$, so the inequality is $x \le 6$.

4. First quadrant constraints

   The shading is entirely in the region where $x\ge0$ and $y\ge0$, so those are included.

Now check the options:

• Option (A): $4x+5y \ge 20$, $3x+10y \le 30$, $x \le 6$, $x,y\ge0$ — matches all three signs.
• Option (B): $x \ge 6$ — wrong, the shading is left of $x=6$.
• Option (C): $4x+5y \le 20$ — wrong, the shading is above that line.
• Option (D): both $4x+5y \le 20$ and $x \ge 6$ — both wrong.

The problem asks you to shade the feasible region for a system of linear inequalities. This is the foundation of linear programming — you’re finding all points $(x,y)$ that satisfy every constraint at once. Each inequality describes a half-plane; the feasible region is where all those half-planes overlap, and since $x,y\ge0$, we only work in the first quadrant.

Let’s build it step by step.

1. Rewrite each inequality as an equation to draw the boundary lines.

   For $6x+4y\le120$, the boundary is $6x+4y=120$. Simplify by dividing by 2: $3x+2y=60$.

   For $3x+10y\le180$, the boundary is $3x+10y=180$.

   The lines $x=0$ and $y=0$ are the axes.

2. Find where each line meets the axes — these are the intercepts.

   For $3x+2y=60$:

   • If $x=0$, then $2y=60 \Rightarrow y=30$. Point: $(0,30)$.
   • If $y=0$, then $3x=60 \Rightarrow x=20$. Point: $(20,0)$. For $3x+10y=180$:
   • If $x=0$, then $10y=180 \Rightarrow y=18$. Point: $(0,18)$.
   • If $y=0$, then $3x=180 \Rightarrow x=60$. Point: $(60,0)$.

3. Determine which side of each line is the feasible half-plane.

   Test the origin $(0,0)$ because it’s easy — but only if it’s not on the line (it isn’t here).

   • For $6x+4y\le120$: at $(0,0)$, $0\le120$ is true. So the region containing the origin is feasible. Shade below the line $3x+2y=60$.
   • For $3x+10y\le180$: at $(0,0)$, $0\le180$ is true. So shade below the line $3x+10y=180$ as well.
   • $x\ge0$ means the region to the right of the $y$-axis.
   • $y\ge0$ means the region above the $x$-axis.

4. Find the intersection point of the two boundary lines — this is a corner of the feasible region.

   Solve:

Subtract the first from the second: $(3x+10y)-(3x+2y)=180-60 \Rightarrow 8y=120 \Rightarrow y=15$.

Substitute into $3x+2(15)=60 \Rightarrow 3x+30=60 \Rightarrow 3x=30 \Rightarrow x=10$.

So the lines intersect at $(10,15)$.

5. Identify all vertices of the feasible region.

   The region is bounded by the axes and the two lines. The vertices are:

   • $(0,0)$ — origin.
   • $(20,0)$ — where $3x+2y=60$ meets the $x$-axis.
   • $(10,15)$ — intersection of the two lines.
   • $(0,18)$ — where $3x+10y=180$ meets the $y$-axis. Note: $(0,30)$ is not a vertex because the line $3x+2y=60$ is cut off by the other constraint before reaching the $y$-axis. Similarly, $(60,0)$ is outside because $3x+10y=180$ is cut off by the first line.

6. Shade the feasible region.

   In a rough figure, draw the axes. Plot the points $(20,0)$, $(10,15)$, and $(0,18)$. Draw the line from $(20,0)$ to $(10,15)$ to $(0,18)$ — this is the boundary of the feasible region. Also draw the axes from $(0,0)$ to these points. The feasible region is the quadrilateral with vertices $(0,0)$, $(20,0)$, $(10,15)$, and $(0,18)$. Shade the interior of this quadrilateral.