Chapter 12 Linear Programming

We are asked to maximise the objective function $Z = 4x + y$ subject to the given constraints.

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To solve this linear programming problem graphically, we first need to find the feasible region determined by the constraints.

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The constraints are:

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We plot the lines corresponding to the equations obtained by changing the inequalities (2) and (3) into equalities:

Line 1: $x + y = 50$

Line 2: $3x + y = 90$

And the lines $x = 0$ (y-axis) and $y = 0$ (x-axis) from constraint (4).

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For Line 1 ($x + y = 50$):

x	y = 50 - x	Point (x, y)
0	50	(0, 50)
50	0	(50, 0)

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For Line 2 ($3x + y = 90$):

x	$y = 90 - 3x$	Point (x, y)
0	90	(0, 90)
30	$90 - 3(30) = 0$	(30, 0)

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Now we determine the feasible region. The inequalities $x \geq 0$ and $y \geq 0$ restrict the region to the first quadrant.

For $x + y \leq 50$, test the origin (0, 0): $0 + 0 \leq 50 \Rightarrow 0 \leq 50$, which is true. So, the region lies towards the origin with respect to the line $x+y=50$.

For $3x + y \leq 90$, test the origin (0, 0): $3(0) + 0 \leq 90 \Rightarrow 0 \leq 90$, which is true. So, the region lies towards the origin with respect to the line $3x+y=90$.

The feasible region is the area in the first quadrant that is below or on the line $x+y=50$ and below or on the line $3x+y=90$. This region is a convex polygon.

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The vertices (corner points) of the feasible region are the points of intersection of the boundary lines. These are:

1. The origin O $(0, 0)$.

2. The intersection of $y=0$ (x-axis) and $3x+y=90$: Substitute $y=0$ into $3x+y=90 \Rightarrow 3x + 0 = 90 \Rightarrow 3x = 90 \Rightarrow x = 30$. Point A $(30, 0)$.

3. The intersection of $x=0$ (y-axis) and $x+y=50$: Substitute $x=0$ into $x+y=50 \Rightarrow 0 + y = 50 \Rightarrow y = 50$. Point C $(0, 50)$.

4. The intersection of $x+y=50$ and $3x+y=90$. We solve the system:

Subtracting equation (i) from equation (ii):

$(3x + y) - (x + y) = 90 - 50$

$2x = 40$

$x = \frac{40}{2} = 20$

Substitute $x=20$ into equation (i):

$20 + y = 50$

$y = 50 - 20$

$y = 30$

Point of intersection is B $(20, 30)$.

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The vertices of the feasible region are O $(0, 0)$, A $(30, 0)$, B $(20, 30)$, and C $(0, 50)$.

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Now, we evaluate the objective function $Z = 4x + y$ at each vertex to find the maximum value.

Vertex (x, y)	$Z = 4x + y$	Value of Z
O (0, 0)	$4(0) + 0$	0
A (30, 0)	$4(30) + 0$	120
B (20, 30)	$4(20) + 30 = 80 + 30$	110
C (0, 50)	$4(0) + 50$	50

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Comparing the values of Z at the vertices, the maximum value is 120.

The maximum value of Z is 120, which occurs at the point $(30, 0)$.

We are asked to minimise the objective function $Z = 200x + 500y$ subject to the given constraints.

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To solve this linear programming problem graphically, we first need to find the feasible region determined by the constraints.

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The constraints are:

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We plot the lines corresponding to the equations obtained by changing the inequalities (2) and (3) into equalities:

Line 1: $x + 2y = 10$

Line 2: $3x + 4y = 24$

And the lines $x = 0$ (y-axis) and $y = 0$ (x-axis) from constraint (4).

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For Line 1 ($x + 2y = 10$):

x	$y = \frac{10-x}{2}$	Point (x, y)
0	$\frac{10-0}{2} = 5$	(0, 5)
10	$\frac{10-10}{2} = 0$	(10, 0)
4	$\frac{10-4}{2} = 3$	(4, 3)

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For Line 2 ($3x + 4y = 24$):

x	$y = \frac{24-3x}{4}$	Point (x, y)
0	$\frac{24-0}{4} = 6$	(0, 6)
8	$\frac{24-3(8)}{4} = 0$	(8, 0)
4	$\frac{24-3(4)}{4} = \frac{12}{4} = 3$	(4, 3)

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Now we determine the feasible region.

The inequalities $x \geq 0$ and $y \geq 0$ restrict the region to the first quadrant.

For $x + 2y \geq 10$, test the origin (0, 0): $0 + 2(0) \geq 10 \Rightarrow 0 \geq 10$, which is false. So, the region is away from the origin relative to the line $x+2y=10$.

For $3x + 4y \leq 24$, test the origin (0, 0): $3(0) + 4(0) \leq 24 \Rightarrow 0 \leq 24$, which is true. So, the region is towards the origin relative to the line $3x+4y=24$.

The feasible region is the set of points $(x, y)$ in the first quadrant that are on or above the line $x + 2y = 10$ and on or below the line $3x + 4y = 24$. This region is a convex polygon.

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The vertices (corner points) of the feasible region are the points where the boundary lines intersect within or on the boundary of the feasible region. These are:

1. Intersection of $x = 0$ (y-axis) and $x + 2y = 10$:

Substitute $x=0$ into $x+2y=10 \Rightarrow 0 + 2y = 10 \Rightarrow y = 5$. Point $(0, 5)$.

So, $(0, 5)$ is a vertex.

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2. Intersection of $x = 0$ (y-axis) and $3x + 4y = 24$:

Substitute $x=0$ into $3x+4y=24 \Rightarrow 3(0) + 4y = 24 \Rightarrow y = 6$. Point $(0, 6)$.

So, $(0, 6)$ is a vertex.

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3. Intersection of $x + 2y = 10$ and $3x + 4y = 24$. We solve the system:

Multiply equation (i) by 2:

Subtract equation (iii) from equation (ii):

$x = 4$

Substitute $x=4$ into equation (i):

$4 + 2y = 10$

$2y = 10 - 4$

$2y = 6$

$y = \frac{6}{2} = 3$

Point of intersection is $(4, 3)$.

So, $(4, 3)$ is a vertex.

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The vertices of the feasible region are $(0, 5)$, $(0, 6)$, and $(4, 3)$.

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Now, we evaluate the objective function $Z = 200x + 500y$ at each vertex to find the minimum value.

Vertex (x, y)	$Z = 200x + 500y$	Value of Z
(0, 5)	$200(0) + 500(5) = 0 + 2500$	2500
(0, 6)	$200(0) + 500(6) = 0 + 3000$	3000
(4, 3)	$200(4) + 500(3) = 800 + 1500$	2300

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Comparing the values of Z at the vertices, the minimum value is 2300.

The minimum value of Z is 2300, which occurs at the point $(4, 3)$.

We are asked to minimise and maximise the objective function $Z = 3x + 9y$ subject to the given constraints.

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To solve this linear programming problem graphically, we first need to find the feasible region determined by the constraints.

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The constraints are:

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We plot the lines corresponding to the equations obtained by changing the inequalities into equalities:

Line 1: $x + 3y = 60$

Line 2: $x + y = 10$

Line 3: $x = y$

And the lines $x = 0$ (y-axis) and $y = 0$ (x-axis) from constraint (5).

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For Line 1 ($x + 3y = 60$):

x	$y = \frac{60-x}{3}$	Point (x, y)
0	$\frac{60-0}{3} = 20$	(0, 20)
60	$\frac{60-60}{3} = 0$	(60, 0)
30	$\frac{60-30}{3} = 10$	(30, 10)

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For Line 2 ($x + y = 10$):

x	y = 10 - x	Point (x, y)
0	10	(0, 10)
10	0	(10, 0)
5	5	(5, 5)

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For Line 3 ($x = y$):

x	y = x	Point (x, y)
0	0	(0, 0)
10	10	(10, 10)
15	15	(15, 15)

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Now we determine the feasible region.

The inequalities $x \geq 0$ and $y \geq 0$ (constraint 5) restrict the region to the first quadrant.

For $x + 3y \leq 60$ (constraint 2), test the origin (0, 0): $0 + 3(0) \leq 60 \Rightarrow 0 \leq 60$, which is true. So, the region lies towards the origin relative to $x+3y=60$.

For $x + y \geq 10$ (constraint 3), test the origin (0, 0): $0 + 0 \geq 10 \Rightarrow 0 \geq 10$, which is false. So, the region lies away from the origin relative to $x+y=10$.

For $x \leq y$ (constraint 4), which is equivalent to $y \geq x$, the region is above or on the line $y=x$. Test point (1,0) below the line: $1 \leq 0$, false. Test point (0,1) above the line: $0 \leq 1$, true. The region is on the side of the line $y=x$ containing the point (0,1).

The feasible region is the set of points $(x, y)$ in the first quadrant that lie on or below the line $x + 3y = 60$, on or above the line $x + y = 10$, and on or above the line $y = x$. This region is a convex polygon.

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The vertices (corner points) of the feasible region are the points where the boundary lines intersect and which are part of the feasible region. These are:

1. Intersection of $x+y=10$ and $x=y$:

Substitute $y=x$ into $x+y=10 \Rightarrow x+x=10 \Rightarrow 2x=10 \Rightarrow x=5$. Since $y=x$, $y=5$. Point (5, 5).

So, (5, 5) is a vertex.

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2. Intersection of $x+3y=60$ and $x=y$:

Substitute $y=x$ into $x+3y=60 \Rightarrow x+3x=60 \Rightarrow 4x=60 \Rightarrow x=15$. Since $y=x$, $y=15$. Point (15, 15).

So, (15, 15) is a vertex.

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3. Intersection of $x+y=10$ and $x=0$ (y-axis):

Substitute $x=0$ into $x+y=10 \Rightarrow 0+y=10 \Rightarrow y=10$. Point (0, 10).

So, (0, 10) is a vertex.

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

Substitute $x=0$ into $x+3y=60 \Rightarrow 0+3y=60 \Rightarrow y=20$. Point (0, 20).

So, (0, 20) is a vertex.

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The vertices of the feasible region are (0, 10), (5, 5), (15, 15), and (0, 20).

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Now, we evaluate the objective function $Z = 3x + 9y$ at each vertex.

Vertex (x, y)	$Z = 3x + 9y$	Value of Z
(0, 10)	$3(0) + 9(10) = 0 + 90$	90
(5, 5)	$3(5) + 9(5) = 15 + 45$	60
(15, 15)	$3(15) + 9(15) = 45 + 135$	180
(0, 20)	$3(0) + 9(20) = 0 + 180$	180

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Comparing the values of Z at the vertices:

The minimum value of Z is 60, which occurs at the point (5, 5).

The maximum value of Z is 180, which occurs at the points (15, 15) and (0, 20).

Since the maximum value occurs at two vertices, it also occurs at every point on the line segment connecting the points (15, 15) and (0, 20).

We are asked to determine the minimum value of the objective function $Z = -50x + 20y$ subject to the given constraints.

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To solve this linear programming problem graphically, we first need to find the feasible region determined by the constraints.

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The constraints are:

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We plot the lines corresponding to the equations obtained by changing the inequalities into equalities:

Line 1: $y = 2x + 5$ (from $2x - y = -5$)

Line 2: $y = 3 - 3x$ (from $3x + y = 3$)

Line 3: $y = \frac{2}{3}x - 4$ (from $2x - 3y = 12$)

And the lines $x = 0$ (y-axis) and $y = 0$ (x-axis) from constraint (5).

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Plotting points for the lines:

For Line 1 ($y = 2x + 5$):

x	y = 2x + 5	Point (x, y)
0	5	(0, 5)
-2.5	0	(-2.5, 0)

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For Line 2 ($y = 3 - 3x$):

x	y = 3 - 3x	Point (x, y)
0	3	(0, 3)
1	0	(1, 0)

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For Line 3 ($y = \frac{2}{3}x - 4$):

x	$y = \frac{2}{3}x - 4$	Point (x, y)
0	-4	(0, -4)
6	0	(6, 0)

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Now we determine the feasible region.

Constraints $x \geq 0, y \geq 0$ restrict the region to the first quadrant.

For $2x - y \geq -5 \implies y \leq 2x + 5$, test origin (0,0): $0 \leq 5$, True. Region is below or on Line 1.

For $3x + y \geq 3$, test origin (0,0): $0 \geq 3$, False. Region is above or on Line 2.

For $2x - 3y \leq 12 \implies y \geq \frac{2}{3}x - 4$, test origin (0,0): $0 \geq -4$, True. Region is above or on Line 3.

The feasible region is the intersection of these regions in the first quadrant. It is bounded by segments of the lines $y=3-3x$, $y=0$, $y=\frac{2}{3}x-4$, and $y=2x+5$. This region is unbounded.

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The vertices (corner points) of the feasible region are the points where the boundary lines intersect within or on the boundary of the feasible region in the first quadrant ($x \geq 0, y \geq 0$).

Let's find the intersection points of the boundary lines that form the vertices of the feasible region:

• Intersection of $x=0$ and $y=3-3x$: $(0, 3)$. This point is a vertex.
• Intersection of $x=0$ and $y=2x+5$: $(0, 5)$. This point is a vertex.
• Intersection of $y=0$ and $y=3-3x$: $0 = 3-3x \implies 3x=3 \implies x=1$. Point $(1, 0)$. This point is a vertex.
• Intersection of $y=0$ and $y=\frac{2}{3}x-4$: $0 = \frac{2}{3}x-4 \implies 4 = \frac{2}{3}x \implies x = 4 \times \frac{3}{2} = 6$. Point $(6, 0)$. This point is a vertex.
• Other intersections ($y=3-3x$ and $y=2x+5$, $y=3-3x$ and $y=\frac{2}{3}x-4$, $y=2x+5$ and $y=\frac{2}{3}x-4$) occur outside the first quadrant, as checked in the original text.

The vertices of the feasible region in the first quadrant are (0, 3), (0, 5), (1, 0), and (6, 0).

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Evaluating the objective function $Z = -50x + 20y$ at each vertex:

Vertex (x, y)	$Z = -50x + 20y$	Value of Z
(0, 3)	$-50(0) + 20(3) = 0 + 60$	60
(0, 5)	$-50(0) + 20(5) = 0 + 100$	100
(1, 0)	$-50(1) + 20(0) = -50 + 0$	-50
(6, 0)	$-50(6) + 20(0) = -300 + 0$	-300

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The feasible region is unbounded. The smallest value of Z at the vertices is -300 at (6, 0).

For an unbounded feasible region, the minimum or maximum value may not exist. We need to check if the objective function can take values smaller than -300 within the feasible region.

Consider the inequality $-50x + 20y < -300$, which represents open half-planes where $Z < -300$. If any point in the feasible region lies in this half-plane, then -300 is not the minimum value.

The line for $Z = -300$ is $-50x + 20y = -300$, or $-5x + 2y = -30$.

Test a point in the feasible region, for example, a point far out along the boundary $y = \frac{2}{3}x - 4$. Let $x = 10$. Then $y = \frac{2}{3}(10) - 4 = \frac{20}{3} - 4 = \frac{8}{3}$. Point $(10, 8/3)$. This point is in the feasible region since $x=10 \geq 0, y=8/3 \geq 0$.

Check constraints: $2(10) - 8/3 = 20 - 8/3 = (60-8)/3 = 52/3 \approx 17.33 \geq -5$ (True). $3(10) + 8/3 = 30 + 8/3 = (90+8)/3 = 98/3 \approx 32.67 \geq 3$ (True). $2(10) - 3(8/3) = 20 - 8 = 12 \leq 12$ (True). So, $(10, 8/3)$ is feasible.

Evaluate Z at $(10, 8/3)$: $Z = -50(10) + 20(8/3) = -500 + 160/3 = (-1500+160)/3 = -1340/3 \approx -446.67$.

Since $-446.67 < -300$, the objective function can take values smaller than -300 within the feasible region.

Also, as we move along the direction in which $x$ increases in the unbounded region, the term $-50x$ becomes a large negative number, and Z decreases without bound.

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Therefore, the minimum value of the objective function $Z = -50x + 20y$ subject to the given constraints does not exist.

The maximum value in this case would occur at (0, 5) or could also not exist, but the question only asks for the minimum.

We are asked to minimise the objective function $Z = 3x + 2y$ subject to the given constraints.

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To solve this linear programming problem graphically, we first need to find the feasible region determined by the constraints.

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The constraints are:

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We plot the lines corresponding to the equations obtained by changing the inequalities (1) and (2) into equalities:

Line 1: $x + y = 8$

Line 2: $3x + 5y = 15$

And the lines $x = 0$ (y-axis) and $y = 0$ (x-axis) from constraint (3).

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For Line 1 ($x + y = 8$):

x	y = 8 - x	Point (x, y)
0	8	(0, 8)
8	0	(8, 0)

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For Line 2 ($3x + 5y = 15$):

x	$y = \frac{15-3x}{5}$	Point (x, y)
0	$\frac{15-0}{5} = 3$	(0, 3)
5	$\frac{15-3(5)}{5} = 0$	(5, 0)

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Now we determine the feasible region.

The inequalities $x \geq 0$ and $y \geq 0$ restrict the region to the first quadrant.

For $x + y \geq 8$, test the origin (0, 0): $0 + 0 \geq 8 \Rightarrow 0 \geq 8$, which is false. So, the region is away from the origin relative to the line $x+y=8$.

For $3x + 5y \leq 15$, test the origin (0, 0): $3(0) + 5(0) \leq 15 \Rightarrow 0 \leq 15$, which is true. So, the region is towards the origin relative to the line $3x+5y=15$.

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We are looking for the set of points $(x, y)$ such that $x \geq 0$, $y \geq 0$, $x+y \geq 8$, and $3x+5y \leq 15$. This is the feasible region.

Let's examine the regions defined by the inequalities in the first quadrant.

The region for $x+y \geq 8$ and $x \geq 0, y \geq 0$ is the area in the first quadrant on or above the line $x+y=8$. This includes points like (0, 8), (8, 0), (10, 0), (0, 10), etc.

The region for $3x+5y \leq 15$ and $x \geq 0, y \geq 0$ is the area in the first quadrant on or below the line $3x+5y=15$. This includes points like (0, 0), (5, 0), (0, 3), (1, 1), etc.

Let's see if there is any overlap between these two regions in the first quadrant.

Consider the points on the line $x+y=8$ in the first quadrant. For these points, $y = 8-x$, where $0 \leq x \leq 8$. Let's check if these points satisfy $3x+5y \leq 15$:

$3x + 5(8-x) \leq 15$

$3x + 40 - 5x \leq 15$

$-2x \leq 15 - 40$

$-2x \leq -25$

Dividing by -2 and reversing the inequality sign:

$x \geq \frac{-25}{-2}$

$x \geq 12.5$

So, for a point on the line $x+y=8$ to also satisfy $3x+5y \leq 15$, its x-coordinate must be $\geq 12.5$. However, the line $x+y=8$ only exists in the first quadrant for $0 \leq x \leq 8$. There is no point on the line $x+y=8$ in the first quadrant where $x \geq 12.5$.

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Alternatively, for any point $(x, y)$ in the first quadrant ($x \ge 0, y \ge 0$) satisfying $x+y \geq 8$, we know $x+y$ is at least 8.

Consider the expression $3x+5y$. Since $y \geq 0$, we have $5y \geq 2y$.

So, $3x+5y \geq 3x+2y = 3(x+y) - x + 2y$. This doesn't seem straightforward.

Let's use a different approach: Since $y \ge 0$, $5y \ge 3y$. Thus, $3x+5y \ge 3x+3y = 3(x+y)$.

If a point $(x,y)$ is in the feasible region, it must satisfy $x+y \geq 8$.

From $x+y \geq 8$ and $x \ge 0, y \ge 0$, the smallest possible value for $3x+5y$ occurs at the points closest to the origin on the line $x+y=8$, which are (8,0) and (0,8). $Z(8,0)=3(8)+2(0)=24$. $Z(0,8)=3(0)+2(8)=16$. Any point $(x,y)$ in the first quadrant with $x+y \ge 8$ will have $3x+5y \ge 16$ (specifically, $3x+5y = 3(x+y)+2y \ge 3(8) + 2y = 24+2y$. Since $y \ge 0$, $3x+5y \ge 24$).

So, any point in the first quadrant satisfying $x+y \geq 8$ must also satisfy $3x+5y \geq 24$.

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However, the second constraint is $3x+5y \leq 15$.

We need to find points $(x,y)$ such that $x \geq 0, y \geq 0$, AND $x+y \geq 8$ AND $3x+5y \leq 15$.

We found that if $x \geq 0, y \geq 0,$ and $x+y \geq 8$, then $3x+5y \geq 24$.

But the requirement is $3x+5y \leq 15$. This is a contradiction ($24 > 15$).

There is no point $(x, y)$ in the first quadrant that can simultaneously satisfy $x+y \geq 8$ and $3x+5y \leq 15$.

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Thus, the set of points satisfying all the constraints is empty.

The feasible region is empty.

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Since there are no points in the feasible region, the linear programming problem has no solution.

Therefore, the objective function $Z = 3x + 2y$ cannot be minimised under the given constraints.

The minimum value of Z does not exist.

We are asked to maximise the objective function $Z = 3x + 4y$ subject to the given constraints.

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The objective function is:

The constraints are:

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To solve this linear programming problem graphically, we first need to find the feasible region determined by the constraints.

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We plot the lines corresponding to the equations obtained by changing the inequalities into equalities:

Line 1: $x + y = 4$

Line 2: $x = 0$ (y-axis)

Line 3: $y = 0$ (x-axis)

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For Line 1 ($x + y = 4$):

x	y = 4 - x	Point (x, y)
0	4	(0, 4)
4	0	(4, 0)

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Now we determine the feasible region.

The constraints $x \geq 0$ and $y \geq 0$ (from inequalities 3 and 4) restrict the region to the first quadrant.

For $x + y \leq 4$ (inequality 2), test the origin (0, 0): $0 + 0 \leq 4 \implies 0 \leq 4$, which is true. So, the region is towards the origin relative to the line $x+y=4$.

The feasible region is the area in the first quadrant bounded by the x-axis ($y=0$), the y-axis ($x=0$), and the line $x + y = 4$. This region is a triangle OAB, where O is the origin, A is the point (4, 0), and B is the point (0, 4).

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The vertices (corner points) of the feasible region are the points of intersection of the boundary lines in the feasible region. These are:

1. The intersection of $x=0$ and $y=0$ is the origin O $(0, 0)$.

2. The intersection of $y=0$ (x-axis) and $x+y=4$: Substitute $y=0$ into $x+y=4 \implies x+0=4 \implies x=4$. Point A $(4, 0)$.

3. The intersection of $x=0$ (y-axis) and $x+y=4$: Substitute $x=0$ into $x+y=4 \implies 0+y=4 \implies y=4$. Point B $(0, 4)$.

The vertices of the feasible region are (0, 0), (4, 0), and (0, 4).

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Now, we evaluate the objective function $Z = 3x + 4y$ at each vertex to find the maximum value.

Vertex (x, y)	$Z = 3x + 4y$	Value of Z
(0, 0)	$3(0) + 4(0) = 0 + 0$	0
(4, 0)	$3(4) + 4(0) = 12 + 0$	12
(0, 4)	$3(0) + 4(4) = 0 + 16$	16

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Comparing the values of Z at the vertices, the maximum value is 16.

The maximum value of Z is 16, which occurs at the point (0, 4).

We are asked to minimise the objective function $Z = -3x + 4y$ subject to the given constraints.

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The objective function is:

The constraints are:

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To solve this linear programming problem graphically, we first need to find the feasible region determined by the constraints.

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We plot the lines corresponding to the equations obtained by changing the inequalities (2) and (3) into equalities:

Line L1: $x + 2y = 8$

Line L2: $3x + 2y = 12$

And the lines $x = 0$ (y-axis) and $y = 0$ (x-axis) from constraint (4).

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For Line L1 ($x + 2y = 8$):

x	$y = \frac{8-x}{2}$	Point (x, y)
0	$\frac{8-0}{2} = 4$	(0, 4)
8	$\frac{8-8}{2} = 0$	(8, 0)
4	$\frac{8-4}{2} = 2$	(4, 2)

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For Line L2 ($3x + 2y = 12$):

x	$y = \frac{12-3x}{2}$	Point (x, y)
0	$\frac{12-0}{2} = 6$	(0, 6)
4	$\frac{12-3(4)}{2} = 0$	(4, 0)
2	$\frac{12-3(2)}{2} = \frac{6}{2} = 3$	(2, 3)

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Now we determine the feasible region.

The inequalities $x \geq 0$ and $y \geq 0$ restrict the region to the first quadrant.

For $x + 2y \leq 8$, test the origin (0, 0): $0 + 2(0) \leq 8 \implies 0 \leq 8$, which is true. So, the region is towards the origin relative to the line $x+2y=8$.

For $3x + 2y \leq 12$, test the origin (0, 0): $3(0) + 2(0) \leq 12 \implies 0 \leq 12$, which is true. So, the region is towards the origin relative to the line $3x+2y=12$.

The feasible region is the area in the first quadrant bounded by the x-axis, the y-axis, and lying below or on both lines $x + 2y = 8$ and $3x + 2y = 12$. This region is a convex polygon.

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The vertices (corner points) of the feasible region are the points of intersection of the boundary lines in the feasible region. These are:

1. The intersection of $x=0$ and $y=0$ is the origin $(0, 0)$.

2. The intersection of $y=0$ (x-axis) and $3x+2y=12$: Substitute $y=0$ into $3x+2y=12 \implies 3x+0=12 \implies x=4$. Point $(4, 0)$.

3. The intersection of $x=0$ (y-axis) and $x+2y=8$: Substitute $x=0$ into $x+2y=8 \implies 0+2y=8 \implies y=4$. Point $(0, 4)$.

4. The intersection of $x+2y=8$ and $3x+2y=12$. We solve the system:

Subtracting equation (i) from equation (ii):

$2x = 4$

$x = \frac{4}{2} = 2$

Substitute $x=2$ into equation (i):

$2 + 2y = 8$

$2y = 8 - 2$

$2y = 6$

$y = \frac{6}{2} = 3$

Point of intersection is $(2, 3)$.

---

The vertices of the feasible region are (0, 0), (4, 0), (2, 3), and (0, 4).

---

Now, we evaluate the objective function $Z = -3x + 4y$ at each vertex to find the minimum value.

Vertex (x, y)	$Z = -3x + 4y$	Value of Z
(0, 0)	$-3(0) + 4(0) = 0 + 0$	0
(4, 0)	$-3(4) + 4(0) = -12 + 0$	-12
(2, 3)	$-3(2) + 4(3) = -6 + 12$	6
(0, 4)	$-3(0) + 4(4) = 0 + 16$	16

---

Comparing the values of Z at the vertices, the minimum value is -12.

The minimum value of Z is -12, which occurs at the point (4, 0).

We are asked to maximise the objective function $Z = 5x + 3y$ subject to the given constraints.

---

The objective function is:

The constraints are:

---

To solve this linear programming problem graphically, we first need to find the feasible region determined by the constraints.

---

We plot the lines corresponding to the equations obtained by changing the inequalities (2) and (3) into equalities:

Line L1: $3x + 5y = 15$

Line L2: $5x + 2y = 10$

And the lines $x = 0$ (y-axis) and $y = 0$ (x-axis) from constraint (4).

---

For Line L1 ($3x + 5y = 15$):

x	$y = \frac{15-3x}{5}$	Point (x, y)
0	$\frac{15-0}{5} = 3$	(0, 3)
5	$\frac{15-3(5)}{5} = 0$	(5, 0)
2.5	$\frac{15-3(2.5)}{5} = \frac{15-7.5}{5} = \frac{7.5}{5} = 1.5$	(2.5, 1.5)

---

For Line L2 ($5x + 2y = 10$):

x	$y = \frac{10-5x}{2}$	Point (x, y)
0	$\frac{10-0}{2} = 5$	(0, 5)
2	$\frac{10-5(2)}{2} = 0$	(2, 0)
1	$\frac{10-5(1)}{2} = \frac{5}{2} = 2.5$	(1, 2.5)

---

Now we determine the feasible region.

The inequalities $x \geq 0$ and $y \geq 0$ restrict the region to the first quadrant.

For $3x + 5y \leq 15$, test the origin (0, 0): $3(0) + 5(0) \leq 15 \implies 0 \leq 15$, which is true. So, the region is towards the origin relative to the line $3x+5y=15$.

For $5x + 2y \leq 10$, test the origin (0, 0): $5(0) + 2(0) \leq 10 \implies 0 \leq 10$, which is true. So, the region is towards the origin relative to the line $5x+2y=10$.

The feasible region is the area in the first quadrant bounded by the x-axis, the y-axis, and lying below or on both lines $3x + 5y = 15$ and $5x + 2y = 10$. This region is a convex polygon.

---

The vertices (corner points) of the feasible region are the points of intersection of the boundary lines in the feasible region. These are:

1. The intersection of $x=0$ and $y=0$ is the origin $(0, 0)$.

2. The intersection of $y=0$ (x-axis) and $5x+2y=10$: Substitute $y=0$ into $5x+2y=10 \implies 5x+0=10 \implies x=2$. Point $(2, 0)$.

So, $(2, 0)$ is a vertex.

---

3. The intersection of $x=0$ (y-axis) and $3x+5y=15$: Substitute $x=0$ into $3x+5y=15 \implies 0+5y=15 \implies y=3$. Point $(0, 3)$.

So, $(0, 3)$ is a vertex.

---

4. The intersection of $3x+5y=15$ and $5x+2y=10$. We solve the system:

To eliminate y, multiply equation (i) by 2 and equation (ii) by 5:

Subtracting equation (iii) from equation (iv):

$19x = 20$

$x = \frac{20}{19}$

Substitute $x = \frac{20}{19}$ into equation (i):

$3\left(\frac{20}{19}\right) + 5y = 15$

$\frac{60}{19} + 5y = 15$

$5y = 15 - \frac{60}{19} = \frac{15 \times 19 - 60}{19} = \frac{285 - 60}{19} = \frac{225}{19}$

$y = \frac{225}{19 \times 5} = \frac{\cancel{225}^{45}}{19 \times \cancel{5}_{1}} = \frac{45}{19}$

Point of intersection is $\left(\frac{20}{19}, \frac{45}{19}\right)$.

So, $\left(\frac{20}{19}, \frac{45}{19}\right)$ is a vertex.

---

The vertices of the feasible region are (0, 0), (2, 0), $\left(\frac{20}{19}, \frac{45}{19}\right)$, and (0, 3).

---

Now, we evaluate the objective function $Z = 5x + 3y$ at each vertex to find the maximum value.

Vertex (x, y)	$Z = 5x + 3y$	Value of Z
(0, 0)	$5(0) + 3(0) = 0 + 0$	0
(2, 0)	$5(2) + 3(0) = 10 + 0$	10
(0, 3)	$5(0) + 3(3) = 0 + 9$	9
$\left(\frac{20}{19}, \frac{45}{19}\right)$	$5\left(\frac{20}{19}\right) + 3\left(\frac{45}{19}\right) = \frac{100}{19} + \frac{135}{19} = \frac{100+135}{19} = \frac{235}{19}$	$\frac{235}{19} \approx 12.37$

---

Comparing the values of Z at the vertices, the maximum value is $\frac{235}{19}$.

The maximum value of Z is $\frac{235}{19}$, which occurs at the point $\left(\frac{20}{19}, \frac{45}{19}\right)$.

We are asked to minimise the objective function $Z = 3x + 5y$ subject to the given constraints.

---

The objective function is:

The constraints are:

---

To solve this linear programming problem graphically, we first need to find the feasible region determined by the constraints.

---

We plot the lines corresponding to the equations obtained by changing the inequalities (2) and (3) into equalities:

Line L1: $x + 3y = 3$

Line L2: $x + y = 2$

And the lines $x = 0$ (y-axis) and $y = 0$ (x-axis) from constraint (4).

---

For Line L1 ($x + 3y = 3$):

x	$y = \frac{3-x}{3}$	Point (x, y)
0	$\frac{3-0}{3} = 1$	(0, 1)
3	$\frac{3-3}{3} = 0$	(3, 0)

---

For Line L2 ($x + y = 2$):

x	y = 2 - x	Point (x, y)
0	2	(0, 2)
2	0	(2, 0)

---

Now we determine the feasible region.

The inequalities $x \geq 0$ and $y \geq 0$ restrict the region to the first quadrant.

For $x + 3y \geq 3$, test the origin (0, 0): $0 + 3(0) \geq 3 \implies 0 \geq 3$, which is false. So, the region is away from the origin relative to the line $x+3y=3$. (i.e., above or on the line).

For $x + y \geq 2$, test the origin (0, 0): $0 + 0 \geq 2 \implies 0 \geq 2$, which is false. So, the region is away from the origin relative to the line $x+y=2$. (i.e., above or on the line).

The feasible region is the area in the first quadrant lying above or on the line $x + 3y = 3$ and above or on the line $x + y = 2$. This region is an unbounded convex region.

---

The vertices (corner points) of the feasible region are the points where the boundary lines intersect and which are part of the feasible region. These are:

1. Intersection of $x=0$ (y-axis) and $x+y=2$: Substitute $x=0$ into $x+y=2 \implies 0+y=2 \implies y=2$. Point $(0, 2)$.

So, $(0, 2)$ is a vertex.

---

2. Intersection of $y=0$ (x-axis) and $x+3y=3$: Substitute $y=0$ into $x+3y=3 \implies x+0=3 \implies x=3$. Point $(3, 0)$.

So, $(3, 0)$ is a vertex.

---

3. Intersection of $x+3y=3$ and $x+y=2$. We solve the system:

Subtracting equation (ii) from equation (i):

$2y = 1$

$y = \frac{1}{2}$

Substitute $y = \frac{1}{2}$ into equation (ii):

$x + \frac{1}{2} = 2$

$x = 2 - \frac{1}{2} = \frac{4-1}{2} = \frac{3}{2}$

Point of intersection is $\left(\frac{3}{2}, \frac{1}{2}\right)$.

So, $\left(\frac{3}{2}, \frac{1}{2}\right)$ is a vertex.

---

The vertices of the feasible region are (3, 0), $\left(\frac{3}{2}, \frac{1}{2}\right)$, and (0, 2).

---

Now, we evaluate the objective function $Z = 3x + 5y$ at each vertex to find the minimum value.

Vertex (x, y)	$Z = 3x + 5y$	Value of Z
(3, 0)	$3(3) + 5(0) = 9 + 0$	9
$\left(\frac{3}{2}, \frac{1}{2}\right)$	$3\left(\frac{3}{2}\right) + 5\left(\frac{1}{2}\right) = \frac{9}{2} + \frac{5}{2} = \frac{14}{2}$	7
(0, 2)	$3(0) + 5(2) = 0 + 10$	10

---

The smallest value of Z at the vertices is 7 at $\left(\frac{3}{2}, \frac{1}{2}\right)$.

The feasible region is unbounded. For an unbounded feasible region, the minimum value may or may not exist. If it exists, it must occur at a vertex.

To verify if 7 is the minimum value, consider the open half-plane defined by $3x + 5y < 7$. If this half-plane contains any point from the feasible region, then 7 is not the minimum value.

The line $3x + 5y = 7$ passes through the vertex $\left(\frac{3}{2}, \frac{1}{2}\right)$. The feasible region lies above or on the lines $x+3y=3$ and $x+y=2$. If we test a point in the region $3x+5y < 7$, e.g., (0, 0), $0 < 7$ (True), but (0,0) is not in the feasible region.

The region $3x+5y < 7$ lies below the line $3x+5y=7$. Visualizing the graph, the feasible region is entirely on or above the line $3x+5y=7$. Therefore, the minimum value exists and is 7.

---

The minimum value of Z is 7, which occurs at the point $\left(\frac{3}{2}, \frac{1}{2}\right)$.

We are asked to maximise the objective function $Z = 3x + 2y$ subject to the given constraints.

---

The objective function is:

The constraints are:

---

To solve this linear programming problem graphically, we first need to find the feasible region determined by the constraints.

---

We plot the lines corresponding to the equations obtained by changing the inequalities (2) and (3) into equalities:

Line L1: $x + 2y = 10$

Line L2: $3x + y = 15$

And the lines $x = 0$ (y-axis) and $y = 0$ (x-axis) from constraint (4).

---

For Line L1 ($x + 2y = 10$):

x	$y = \frac{10-x}{2}$	Point (x, y)
0	$\frac{10-0}{2} = 5$	(0, 5)
10	$\frac{10-10}{2} = 0$	(10, 0)
4	$\frac{10-4}{2} = 3$	(4, 3)

---

For Line L2 ($3x + y = 15$):

x	y = 15 - 3x	Point (x, y)
0	15	(0, 15)
5	$15 - 3(5) = 0$	(5, 0)
4	$15 - 3(4) = 3$	(4, 3)

---

Now we determine the feasible region.

The inequalities $x \geq 0$ and $y \geq 0$ restrict the region to the first quadrant.

For $x + 2y \leq 10$, test the origin (0, 0): $0 + 2(0) \leq 10 \implies 0 \leq 10$, which is true. So, the region is towards the origin relative to the line $x+2y=10$.

For $3x + y \leq 15$, test the origin (0, 0): $3(0) + 0 \leq 15 \implies 0 \leq 15$, which is true. So, the region is towards the origin relative to the line $3x+y=15$.

The feasible region is the area in the first quadrant bounded by the x-axis, the y-axis, and lying below or on both lines $x + 2y = 10$ and $3x + y = 15$. This region is a convex polygon.

---

The vertices (corner points) of the feasible region are the points of intersection of the boundary lines in the feasible region. These are:

1. The intersection of $x=0$ and $y=0$ is the origin $(0, 0)$.

2. The intersection of $y=0$ (x-axis) and $3x+y=15$: Substitute $y=0$ into $3x+y=15 \implies 3x+0=15 \implies x=5$. Point $(5, 0)$.

So, $(5, 0)$ is a vertex.

---

3. The intersection of $x=0$ (y-axis) and $x+2y=10$: Substitute $x=0$ into $x+2y=10 \implies 0+2y=10 \implies y=5$. Point $(0, 5)$.

So, $(0, 5)$ is a vertex.

---

4. The intersection of $x+2y=10$ and $3x+y=15$. We solve the system:

From equation (ii), we can write $y = 15 - 3x$. Substitute this into equation (i):

$x + 2(15 - 3x) = 10$

$x + 30 - 6x = 10$

$-5x = 10 - 30$

$-5x = -20$

$x = \frac{-20}{-5} = 4$

Substitute $x=4$ into the expression for y: $y = 15 - 3(4) = 15 - 12 = 3$.

Point of intersection is $(4, 3)$.

So, $(4, 3)$ is a vertex.

---

The vertices of the feasible region are (0, 0), (5, 0), (4, 3), and (0, 5).

---

Now, we evaluate the objective function $Z = 3x + 2y$ at each vertex to find the maximum value.

Vertex (x, y)	$Z = 3x + 2y$	Value of Z
(0, 0)	$3(0) + 2(0) = 0 + 0$	0
(5, 0)	$3(5) + 2(0) = 15 + 0$	15
(4, 3)	$3(4) + 2(3) = 12 + 6$	18
(0, 5)	$3(0) + 2(5) = 0 + 10$	10

---

Comparing the values of Z at the vertices, the maximum value is 18.

The maximum value of Z is 18, which occurs at the point (4, 3).

We are asked to minimise the objective function $Z = x + 2y$ subject to the given constraints.

---

The objective function is:

The constraints are:

---

To solve this linear programming problem graphically, we first need to find the feasible region determined by the constraints.

---

We plot the lines corresponding to the equations obtained by changing the inequalities (2) and (3) into equalities:

Line L1: $2x + y = 3$

Line L2: $x + 2y = 6$

And the lines $x = 0$ (y-axis) and $y = 0$ (x-axis) from constraint (4).

---

For Line L1 ($2x + y = 3$):

x	y = 3 - 2x	Point (x, y)
0	3	(0, 3)
1.5	$3 - 2(1.5) = 0$	(1.5, 0)
1	$3 - 2(1) = 1$	(1, 1)

---

For Line L2 ($x + 2y = 6$):

x	$y = \frac{6-x}{2}$	Point (x, y)
0	$\frac{6-0}{2} = 3$	(0, 3)
6	$\frac{6-6}{2} = 0$	(6, 0)
4	$\frac{6-4}{2} = 1$	(4, 1)

---

Now we determine the feasible region.

The inequalities $x \geq 0$ and $y \geq 0$ restrict the region to the first quadrant.

For $2x + y \geq 3$, test the origin (0, 0): $2(0) + 0 \geq 3 \implies 0 \geq 3$, which is false. So, the region is away from the origin relative to the line $2x+y=3$. (i.e., above or on the line).

For $x + 2y \geq 6$, test the origin (0, 0): $0 + 2(0) \geq 6 \implies 0 \geq 6$, which is false. So, the region is away from the origin relative to the line $x+2y=6$. (i.e., above or on the line).

The feasible region is the set of points $(x, y)$ in the first quadrant that lie on or above the line $2x + y = 3$ AND on or above the line $x + 2y = 6$. This region is an unbounded convex region.

---

The vertices (corner points) of the feasible region are the points where the boundary lines intersect and which are part of the feasible region. These are:

1. Intersection of $x=0$ (y-axis) and $x+2y=6$: Substitute $x=0$ into $x+2y=6 \implies 0+2y=6 \implies y=3$. Point $(0, 3)$.

So, $(0, 3)$ is a vertex.

---

2. Intersection of $y=0$ (x-axis) and $x+2y=6$: Substitute $y=0$ into $x+2y=6 \implies x+0=6 \implies x=6$. Point $(6, 0)$.

So, $(6, 0)$ is a vertex.

---

3. Intersection of $2x+y=3$ and $x+2y=6$. We solve the system:

From (i), $y = 3 - 2x$. Substitute this into (ii):

$x + 2(3 - 2x) = 6$

$x + 6 - 4x = 6$

$-3x = 0$

$x = 0$

Substitute $x=0$ into $y = 3 - 2x$: $y = 3 - 2(0) = 3$. Point of intersection is $(0, 3)$. This is the same as vertex 1.

The intersection point of the other two lines, $y=0$ and $2x+y=3$, is (1.5, 0), but this point does not satisfy $x+2y \ge 6$ ($1.5+0 \ge 6$ is False), so it is not a vertex of the feasible region.

---

The feasible region is the unbounded region in the first quadrant bounded by the line segment connecting (0, 3) and (6, 0), and the rays extending upwards from (0, 3) along the y-axis and rightwards from (6, 0) along the x-axis. The vertices are (0, 3) and (6, 0).

---

Now, we evaluate the objective function $Z = x + 2y$ at each vertex.

Vertex (x, y)	$Z = x + 2y$	Value of Z
(0, 3)	$0 + 2(3)$	6
(6, 0)	$6 + 2(0)$	6

---

Both vertices give the same value of Z, which is 6.

The feasible region is unbounded. We need to check if the minimum value exists. The objective function is $Z = x + 2y$. The lines $x+2y=k$ are the level sets of Z.

The constraint $x+2y \ge 6$ directly means that any point $(x,y)$ in the feasible region must satisfy $x+2y \ge 6$. This tells us the minimum value of the expression $x+2y$ in the feasible region is at least 6.

Since the line $x+2y=6$ itself forms part of the boundary of the feasible region (the segment connecting (0,3) and (6,0)), the value $Z=6$ is achieved for all points on this segment within the feasible region.

To confirm that 6 is the minimum, consider the open half-plane $x+2y < 6$. This region is below the line $x+2y=6$. Since the feasible region is entirely on or above the line $x+2y=6$, there are no points in the feasible region that satisfy $x+2y < 6$. Thus, the minimum value exists and is 6.

The minimum value of Z is 6, which occurs at every point on the line segment connecting the points (0, 3) and (6, 0).

The question asks to "Show that the minimum of Z occurs at more than two points". Since a line segment consists of infinitely many points, the minimum occurs at more than two points (in fact, infinitely many points) on this segment.

---

The minimum value of Z is 6, which occurs at every point on the line segment connecting (0, 3) and (6, 0).

We are asked to minimise and maximise the objective function $Z = 5x + 10y$ subject to the given constraints.

---

The objective function is:

The constraints are:

---

To solve this linear programming problem graphically, we first need to find the feasible region determined by the constraints.

---

We plot the lines corresponding to the equations obtained by changing the inequalities into equalities:

Line L1: $x + 2y = 120$

Line L2: $x + y = 60$

Line L3: $y = \frac{1}{2}x$ (or $x - 2y = 0$)

And the lines $x = 0$ (y-axis) and $y = 0$ (x-axis) from constraint (5).

---

Plotting points for the lines:

For Line L1 ($x + 2y = 120$):

x	$y = \frac{120-x}{2}$	Point (x, y)
0	$\frac{120-0}{2} = 60$	(0, 60)
120	$\frac{120-120}{2} = 0$	(120, 0)
60	$\frac{120-60}{2} = 30$	(60, 30)

---

For Line L2 ($x + y = 60$):

x	$y = 60-x$	Point (x, y)
0	60	(0, 60)
60	0	(60, 0)
40	$60-40 = 20$	(40, 20)

---

For Line L3 ($y = \frac{1}{2}x$):

x	$y = \frac{1}{2}x$	Point (x, y)
0	0	(0, 0)
60	$\frac{1}{2}(60) = 30$	(60, 30)
40	$\frac{1}{2}(40) = 20$	(40, 20)

---

Now we determine the feasible region.

The inequalities $x \geq 0$ and $y \geq 0$ (constraint 5) restrict the region to the first quadrant.

For $x + 2y \leq 120$ (constraint 2), test the origin (0, 0): $0 + 2(0) \leq 120 \implies 0 \leq 120$, which is true. So, the region lies towards the origin relative to L1.

For $x + y \geq 60$ (constraint 3), test the origin (0, 0): $0 + 0 \geq 60 \implies 0 \geq 60$, which is false. So, the region lies away from the origin relative to L2.

For $y \leq \frac{1}{2}x$ (constraint 4), or $x - 2y \geq 0$. Test a point not on the line $y=x/2$, e.g., (1,0): $1 - 2(0) = 1 \ge 0$, which is true. So, the region is towards the side of L3 containing (1,0), which is below or on the line $y=x/2$ in the first quadrant.

The feasible region is the set of points $(x, y)$ in the first quadrant that lie on or below the line $x+2y=120$, on or above the line $x+y=60$, and on or below the line $y=x/2$. This region is a convex polygon.

---

The vertices (corner points) of this feasible region are the points of intersection of the boundary lines that satisfy all constraints. These are:

• Intersection of $y=0$ (x-axis) and $x+y=60$: Substitute $y=0$ into $x+y=60 \implies x=60$. Point $(60, 0)$. Check other constraints: $x \ge 0, y \ge 0$ (True), $x+2y \le 120 \implies 60+0 \le 120$ (True), $y \le x/2 \implies 0 \le 60/2=30$ (True). Vertex.
• Intersection of $y=0$ (x-axis) and $x+2y=120$: Substitute $y=0$ into $x+2y=120 \implies x=120$. Point $(120, 0)$. Check other constraints: $x \ge 0, y \ge 0$ (True), $x+y \ge 60 \implies 120+0 \ge 60$ (True), $y \le x/2 \implies 0 \le 120/2=60$ (True). Vertex.
• Intersection of $x+y=60$ and $y=x/2$: Substitute $y=x/2$ into $x+y=60 \implies x+x/2=60 \implies 3x/2=60 \implies x = 60 \times \frac{2}{3} = 40$. Then $y = 40/2 = 20$. Point $(40, 20)$. Check other constraints: $x \ge 0, y \ge 0$ (True), $x+2y \le 120 \implies 40+2(20)=40+40=80 \le 120$ (True). Vertex.
• Intersection of $x+2y=120$ and $y=x/2$: Substitute $y=x/2$ into $x+2y=120 \implies x+2(x/2)=120 \implies x+x=120 \implies 2x=120 \implies x=60$. Then $y=60/2=30$. Point $(60, 30)$. Check other constraints: $x \ge 0, y \ge 0$ (True), $x+y \ge 60 \implies 60+30=90 \ge 60$ (True). Vertex.

The vertices of the feasible region are (60, 0), (120, 0), (40, 20), and (60, 30).

---

Now, we evaluate the objective function $Z = 5x + 10y$ at each vertex.

Vertex (x, y)	$Z = 5x + 10y$	Value of Z
(60, 0)	$5(60) + 10(0) = 300 + 0$	300
(120, 0)	$5(120) + 10(0) = 600 + 0$	600
(40, 20)	$5(40) + 10(20) = 200 + 200$	400
(60, 30)	$5(60) + 10(30) = 300 + 300$	600

---

Comparing the values of Z at the vertices:

The minimum value of Z is 300, which occurs at the vertex (60, 0).

The maximum value of Z is 600, which occurs at vertices (120, 0) and (60, 30).

Since the maximum value occurs at two vertices, it also occurs at every point on the line segment connecting the points (120, 0) and (60, 30). This line segment lies on the line $x+2y=120$, and for any point $(x,y)$ on this segment, $Z = 5(x+2y) = 5(120) = 600$.

We are asked to minimise and maximise the objective function $Z = x + 2y$ subject to the given constraints.

---

The objective function is:

The constraints are:

---

To solve this linear programming problem graphically, we first need to find the feasible region determined by the constraints.

---

We plot the lines corresponding to the equations obtained by changing the inequalities into equalities:

Line L1: $x + 2y = 100$

Line L2: $y = 2x$ (or $2x - y = 0$)

Line L3: $2x + y = 200$

And the lines $x = 0$ (y-axis) and $y = 0$ (x-axis) from constraint (5).

---

Plotting points for the lines:

For Line L1 ($x + 2y = 100$):

x	$y = \frac{100-x}{2}$	Point (x, y)
0	50	(0, 50)
100	0	(100, 0)
50	25	(50, 25)

---

For Line L2 ($y = 2x$):

x	y = 2x	Point (x, y)
0	0	(0, 0)
10	20	(10, 20)
50	100	(50, 100)

---

For Line L3 ($2x + y = 200$):

x	y = 200 - 2x	Point (x, y)
0	200	(0, 200)
100	0	(100, 0)
50	100	(50, 100)

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Now we determine the feasible region.

Constraints $x \geq 0$ and $y \geq 0$ restrict the region to the first quadrant.

For $x + 2y \geq 100$, test (0, 0): $0 + 2(0) \geq 100 \implies 0 \geq 100$ (False). Region is away from the origin w.r.t. L1 (i.e., above or on the line).

For $y \geq 2x$ (or $2x-y \leq 0$), test a point (1, 0) not on the line $y=2x$: $0 \geq 2(1) \implies 0 \geq 2$ (False). Test a point (0, 1): $1 \geq 2(0) \implies 1 \geq 0$ (True). Region is above or on Line L2 ($y=2x$).

For $2x + y \leq 200$, test (0, 0): $2(0) + 0 \leq 200 \implies 0 \leq 200$ (True). Region is towards the origin w.r.t. L3 (i.e., below or on the line).

The feasible region is the set of points $(x, y)$ in the first quadrant that are on or above $x+2y=100$, on or above $y=2x$, and on or below $2x+y=200$. This region is a convex polygon.

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The vertices (corner points) of the feasible region are the points where the boundary lines intersect within the feasible region. These are:

• Intersection of $x=0$ (y-axis) and $x+2y=100$: Substitute $x=0$ into $0 + 2y = 100 \implies y = 50$. Point (0, 50).

  Check constraints $y \ge 2x$ and $2x+y \le 200$ at (0, 50):

  $50 \ge 2(0) \implies 50 \ge 0$

  (True)

  $2(0) + 50 = 50$

  $50 \leq 200$

  (True)

  So, (0, 50) is a vertex.
• Intersection of $x=0$ (y-axis) and $2x+y=200$: Substitute $x=0$ into $2(0) + y = 200 \implies y = 200$. Point (0, 200).

  Check constraints $x+2y \ge 100$ and $y \ge 2x$ at (0, 200):

  $0 + 2(200) = 400$

  $400 \geq 100$

  (True)

  $200 \ge 2(0) \implies 200 \ge 0$

  (True)

  So, (0, 200) is a vertex.
• Intersection of $x+2y=100$ and $y=2x$: Substitute $y=2x$ into $x+2y=100 \implies x + 2(2x) = 100 \implies x + 4x = 100 \implies 5x = 100 \implies x=20$. Then $y=2(20)=40$. Point (20, 40).

  Check constraint $2x+y \le 200$ at (20, 40):

  $2(20) + 40 = 40 + 40 = 80$

  $80 \leq 200$

  (True)

  So, (20, 40) is a vertex.
• Intersection of $y=2x$ and $2x+y=200$: Substitute $y=2x$ into $2x+y=200 \implies 2x + 2x = 200 \implies 4x = 200 \implies x=50$. Then $y=2(50)=100$. Point (50, 100).

  Check constraint $x+2y \ge 100$ at (50, 100):

  $50 + 2(100) = 50 + 200 = 250$

  $250 \geq 100$

  (True)

  So, (50, 100) is a vertex.

The vertices of the feasible region are (0, 50), (0, 200), (20, 40), and (50, 100).

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Now, we evaluate the objective function $Z = x + 2y$ at each vertex.

Vertex (x, y)	$Z = x + 2y$	Value of Z
(0, 50)	$0 + 2(50) = 100$	100
(0, 200)	$0 + 2(200) = 400$	400
(20, 40)	$20 + 2(40) = 20 + 80$	100
(50, 100)	$50 + 2(100) = 50 + 200$	250

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Comparing the values of Z at the vertices:

The minimum value of Z is 100.

The maximum value of Z is 400.

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The minimum value of Z is 100, which occurs at vertices (0, 50) and (20, 40). The equation $Z=100$ gives $x+2y=100$, which is exactly the line L1. The line segment connecting these two vertices is part of the boundary of the feasible region. Any point $(x, y)$ on this segment satisfies $x+2y=100$ and lies within the feasible region. Therefore, the minimum value occurs at every point on the line segment connecting (0, 50) and (20, 40).

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The maximum value of Z is 400, which occurs at the vertex (0, 200).

We are asked to minimise and maximise the objective function $Z = -x + 2y$ subject to the given constraints.

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The objective function is:

The constraints are:

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To solve this linear programming problem graphically, we first need to find the feasible region determined by the constraints.

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We plot the lines corresponding to the equations obtained by changing the inequalities into equalities:

Line L1: $x = 3$

Line L2: $x + y = 5$

Line L3: $x + 2y = 6$

Line L4: $y = 0$ (x-axis)

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Plotting points for the lines:

For Line L2 ($x + y = 5$): (0, 5), (5, 0), (3, 2), (4, 1)

For Line L3 ($x + 2y = 6$): (0, 3), (6, 0), (2, 2), (4, 1), (3, 1.5)

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Now we determine the feasible region.

Constraint $x \geq 3$ restricts the region to the right of or on the vertical line $x=3$.

Constraint $y \geq 0$ restricts the region to be on or above the x-axis.

For $x + y \geq 5$, test a point not on the line, e.g., (3, 0): $3 + 0 = 3 \geq 5$, which is false. So, the region is away from (3,0) relative to L2 (i.e., above or on the line $x+y=5$).

For $x + 2y \geq 6$, test a point not on the line, e.g., (3, 0): $3 + 2(0) = 3 \geq 6$, which is false. So, the region is away from (3,0) relative to L3 (i.e., above or on the line $x+2y=6$).

The feasible region is the set of points $(x, y)$ in the first quadrant satisfying all constraints. It is the area on or to the right of $x=3$, on or above $y=0$, on or above $x+y=5$, and on or above $x+2y=6$. This region is an unbounded convex region.

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The vertices (corner points) of the feasible region are the points where the boundary lines intersect within the feasible region. These are:

• Intersection of $x=3$ and $x+y=5$: Substitute $x=3$ into $3+y=5 \implies y=2$. Point (3, 2).

  Check constraints $x+2y \ge 6$ and $y \ge 0$ at (3, 2):

  $3 + 2(2) = 7 \ge 6$

  (True)

  $2 \ge 0$

  (True)

  So, (3, 2) is a vertex.
• Intersection of $x+y=5$ and $x+2y=6$: Subtracting the first equation from the second gives $(x+2y)-(x+y) = 6-5 \implies y=1$. Substitute $y=1$ into $x+y=5 \implies x+1=5 \implies x=4$. Point (4, 1).

  Check constraints $x \ge 3$ and $y \ge 0$ at (4, 1):

  $4 \ge 3$

  (True)

  $1 \ge 0$

  (True)

  So, (4, 1) is a vertex.
• Intersection of $y=0$ (x-axis) and $x+2y=6$: Substitute $y=0$ into $x+2(0)=6 \implies x=6$. Point (6, 0).

  Check constraints $x \ge 3$ and $x+y \ge 5$ at (6, 0):

  $6 \ge 3$

  (True)

  $6+0 = 6 \ge 5$

  (True)

  So, (6, 0) is a vertex.

Other potential intersections, like $x=3$ and $x+2y=6$ (giving (3, 1.5)), or $y=0$ and $x+y=5$ (giving (5,0)), do not satisfy all constraints and hence are not vertices of the feasible region.

The vertices of the feasible region are (3, 2), (4, 1), and (6, 0).

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Now, we evaluate the objective function $Z = -x + 2y$ at each vertex.

Vertex (x, y)	$Z = -x + 2y$	Value of Z
(3, 2)	$-3 + 2(2) = -3 + 4$	1
(4, 1)	$-4 + 2(1) = -4 + 2$	-2
(6, 0)	$-6 + 2(0) = -6 + 0$	-6

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Comparing the values of Z at the vertices, the smallest value is -6 and the largest value is 1.

The feasible region is unbounded. For an unbounded feasible region, the minimum or maximum value may not exist. We need to check if the objective function is bounded over this region.

Consider the ray from vertex (3, 2) along the line $x=3$ for $y \geq 2$. Any point on this ray is of the form $(3, y)$ with $y \geq 2$. The value of the objective function is $Z(3, y) = -3 + 2y$. As $y$ increases (moves upwards in the feasible region), $Z$ increases without bound ($Z \to \infty$). Thus, Z is unbounded above in the feasible region.

Consider the ray from vertex (6, 0) along the line $y=0$ for $x \geq 6$. Any point on this ray is of the form $(x, 0)$ with $x \geq 6$. The value of the objective function is $Z(x, 0) = -x + 2(0) = -x$. As $x$ increases (moves rightwards in the feasible region), $Z$ decreases without bound ($Z \to -\infty$). Thus, Z is unbounded below in the feasible region.

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Since the objective function Z can take arbitrarily large positive values and arbitrarily large negative values within the feasible region, the maximum and minimum values of Z do not exist.

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The minimum value of Z does not exist.

The maximum value of Z does not exist.

We are asked to maximise the objective function $Z = x + y$ subject to the given constraints.

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The objective function is:

The constraints are:

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To solve this linear programming problem graphically, we first need to find the feasible region determined by the constraints.

---

We plot the lines corresponding to the equations obtained by changing the inequalities into equalities:

Line L1: $y = x + 1$

Line L2: $y = x$

And the lines $x = 0$ (y-axis) and $y = 0$ (x-axis) from constraint (4).

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Plotting points for the lines:

For Line L1 ($y = x + 1$):

x	y = x + 1	Point (x, y)
0	1	(0, 1)
1	2	(1, 2)

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For Line L2 ($y = x$):

x	y = x	Point (x, y)
0	0	(0, 0)
1	1	(1, 1)

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Now we determine the feasible region.

Constraints $x \geq 0$ and $y \geq 0$ restrict the region to the first quadrant.

For $y \geq x + 1$ (Constraint 2): Test (0,0): $0 \ge 0+1 \implies 0 \ge 1$ (False). The region is on or above the line $y = x + 1$.

For $y \leq x$ (Constraint 3): Test (0,1): $1 \le 0$ (False). Test (1,0): $0 \le 1$ (True). The region is on or below the line $y = x$.

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We are looking for points $(x, y)$ such that $x \geq 0$, $y \geq 0$, $y \geq x + 1$, and $y \leq x$.

Consider the two inequalities $y \geq x + 1$ and $y \leq x$.

From $y \geq x + 1$, we know that the value of $y$ must be greater than or equal to $x+1$. Since $x+1$ is always greater than $x$ (for any real $x$), this implies $y > x$.

From $y \leq x$, we know that the value of $y$ must be less than or equal to $x$.

So, we are looking for points $(x, y)$ such that $y > x$ AND $y \leq x$. These two conditions are contradictory; there is no number that is simultaneously strictly greater than $x$ and less than or equal to $x$.

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Therefore, there is no point $(x, y)$ that can satisfy both $y \geq x + 1$ and $y \leq x$ simultaneously.

Since there are no points satisfying these two inequalities, there are no points that satisfy all the given constraints, including $x \geq 0$ and $y \geq 0$.

The set of points satisfying all the constraints is empty.

The feasible region is empty.

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Since the feasible region is empty, there are no points $(x, y)$ that satisfy all the constraints.

A linear programming problem with an empty feasible region has no solution.

Therefore, the objective function $Z = x + y$ cannot be maximised (or minimised) subject to these constraints.

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The maximum value of Z does not exist.