Question:

Maximise Z=−x+2y, subject to the constraints: x≥3,x+y≥5,x+2y≥6,y≥0

Answer:

1. Rewrite the objective function as Z=-x+2y

2. Identify the constraints: x≥3,x+y≥5,x+2y≥6,y≥0

3. Graph the constraints on a coordinate plane

4. Find the feasible region by shading the areas that satisfy the constraints

5. Find the coordinates of the vertices of the feasible region

6. Calculate the value of Z at each vertex

7. Choose the vertex with the highest value of Z as the optimal solution

Question:

Minimise Z=x+2y subject to 2x+y≥3,x+2y≥6,x,y≥0

Answer:

Step 1: Rewrite the objective function and constraints in standard form:

Minimise Z=x+2y

Subject to 2x+y-3≥0 x+2y-6≥0 x≥0 y≥0

Step 2: Introduce slack variables s1 and s2:

Minimise Z=x+2y+0s1+0s2

Subject to 2x+y+(-3)s1≥0 x+2y+(-6)s2≥0 x+0s1+0s2≥0 y+0s1+0s2≥0

Step 3: Introduce the artificial variables a1 and a2:

Minimise Z=x+2y+0s1+0s2-a1-a2

Subject to 2x+y+(-3)s1+a1≥0 x+2y+(-6)s2+a2≥0 x+0s1+0s2-a1≥0 y+0s1+0s2-a2≥0 a1,a2≥0

Question:

Minimize Z=−3x+4y, subject to x+2y≤8, 3x+2y≤12, x≥0, y≥0.

Answer:

Step 1: Write the given constraint equations in standard form.

x + 2y ≤ 8 —————> -x - 2y ≥ -8

3x + 2y ≤ 12 —————> -3x - 2y ≥ -12

Step 2: Write the objective function in standard form.

Z = -3x + 4y —————> -Z = 3x - 4y

Step 3: Write the augmented matrix.

[ -1 -2 | -8 ] [ -3 -2 | -12 ] [ 3 4 | 0 ]

Step 4: Use Gaussian elimination to solve the system.

[ -1 -2 | -8 ] —> [ -1 -2 | -8 ] —> [ 1 0 | 4 ] [ 0 0 | 0 ] —> [ -3 -2 | -12 ] —> [ 0 1 | 4 ] [ 3 4 | 0 ] —> [ 0 0 | 0 ]

Step 5: The solution is x = 4, y = 4, and Z = -12.

Question:

Minimise and Maximise of Z=5x+10y subject to x+2y≤120,x+y≥60,x−2y≥0,x,y≥0

Answer:

Step 1: Write the objective function as an equation: Z = 5x + 10y

Step 2: Write the constraints as equations: x + 2y ≤ 120 x + y ≥ 60 x − 2y ≥ 0 x ≥ 0 y ≥ 0

Step 3: Graph the constraints on a coordinate plane.

Step 4: Find the vertices of the feasible region. Vertex 1: (0, 60) Vertex 2: (60, 0) Vertex 3: (120, 0)

Step 5: Calculate the value of the objective function at each vertex. Vertex 1: Z = 300 Vertex 2: Z = 300 Vertex 3: Z = 600

Step 6: Determine the minimum and maximum values of the objective function. Minimum Value: Z = 300 Maximum Value: Z = 600

Question:

Minimise and Maximise Z=x+2y subject to x+2y≥100,2x−y≤0,2x+y≤200;x,y≥0

Answer:

Step 1: Write the objective function Z=x+2y

Step 2: Write the constraints: x+2y≥100,2x−y≤0,2x+y≤200;x,y≥0

Step 3: Minimise Z by solving the system of equations:

x+2y≥100 2x−y≤0 2x+y≤200 x,y≥0

Step 4: Use the method of substitution to solve the system of equations:

Substitute x=200-2y in the first equation: 200-2y+2y≥100 200≥100

Therefore, the constraint is satisfied.

Step 5: Substitute x=200-2y in the second equation: 2(200-2y)-y≤0 400-4y-y≤0 400-5y≤0

Therefore, the constraint is satisfied when 5y≤400 or y≤80.

Step 6: Substitute x=200-2y in the third equation: 2(200-2y)+y≤200 400-2y+y≤200 400-y≤200

Therefore, the constraint is satisfied when y≤200.

Step 7: Finally, the constraints are satisfied when 0≤y≤80 and 0≤x≤200.

Step 8: To minimise Z, substitute x=200-2y in the objective function: Z=200-2y+2y Z=200

Therefore, the minimum value of Z is 200.

Step 9: To maximise Z, substitute x=200-2y in the objective function: Z=200-2y+2y Z=400

Therefore, the maximum value of Z is 400.

Question:

Minimise Z=3x+5y such that x+3y≥3,x+y≥2,x,y≥0

Answer:

1. Rewrite the given constraints in standard form: x + 3y ≥ 3 → -x - 3y ≤ -3 x + y ≥ 2 → -x - y ≤ -2

2. Write the objective function in the form of a maximisation problem (by multiplying the coefficients of x and y by -1): Maximise Z = -3x - 5y

3. Set up the initial simplex tableau:

| Z | x | y | s1 | s2 | b | | -3 | -1 | -3 | 1 | 0 | -3 | | -5 | -1 | -1 | 0 | 1 | -2 |

4. Select a pivot element: Select the pivot element from the bottom row and the leftmost column (in this case, -3 from the Z column).

5. Perform row operations to create the new tableau:

| Z | x | y | s1 | s2 | b | | 1 | 1/3 | 1 | -1/3 | 0 | 1 | | 0 | 2/3 | 2 | 1/3 | 1 | 2 |

6. Check for optimality: The optimal solution is x = 1, y = 2.

Question:

Maximize Z=5x+3y subject to 3x+5y≤15,5x+2y≤10,x≥0,y≥0

Answer:

Step 1: Identify the objective function to be maximized:

Z = 5x + 3y

Step 2: Identify the constraints:

3x + 5y ≤ 15 5x + 2y ≤ 10 x ≥ 0 y ≥ 0

Step 3: Rewrite the constraints in standard form:

3x + 5y ≤ 15 -3x - 5y ≥ -15

5x + 2y ≤ 10 -5x - 2y ≥ -10

x ≥ 0 -x ≤ 0

y ≥ 0 -y ≤ 0

Step 4: Set up the initial tableau:

x y s1 s2 s3 s4 Z

Step 5: Add slack variables to the constraints:

x y s1 s2 s3 s4 Z 3 5 1 0 0 0 0 5 2 0 1 0 0 0 -1 0 0 0 1 0 0 0 -1 0 0 0 1 0

Step 6: Introduce the objective function:

x y s1 s2 s3 s4 Z 3 5 1 0 0 0 0 5 2 0 1 0 0 0 -1 0 0 0 1 0 0 0 -1 0 0 0 1 0 0 0 0 0 0 0 5

Step 7: Perform row operations to create a basic feasible solution:

x y s1 s2 s3 s4 Z 1 3 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 5

Step 8: Check for optimality:

The solution is optimal since all the coefficients in the objective row are zero except for the Z column.

Question:

Maximise Z=x+y, subject to x−y≤−1,x−y≥0,x,y≥0

Answer:

1. Rewrite the given constraint equations in the standard form:

x - y ≤ -1 -x + y ≥ 0 x ≥ 0 y ≥ 0

2. Construct the feasible region:

The feasible region is the set of all points (x, y) that satisfies the constraint equations.

3. Find the corner points of the feasible region:

The corner points of the feasible region are (0, 0), (0, -1), (1, 0), and (1, -1).

4. Calculate the value of the objective function at each corner point:

At (0, 0): Z = 0 + 0 = 0 At (0, -1): Z = 0 + (-1) = -1 At (1, 0): Z = 1 + 0 = 1 At (1, -1): Z = 1 + (-1) = 0

5. Select the corner point which maximises the objective function:

The corner point which maximises the objective function is (1, 0).

Hence, the maximum value of Z is 1.

Question:

Maximise Z=3x+2y subject to x+2y≤10,3x+y≤15,x,y≥0

Answer:

Step 1: Rewrite the objective function and the constraints in standard form.

Objective Function: Maximise Z = 3x + 2y

Constraints: x + 2y ≤ 10 3x + y ≤ 15 x, y ≥ 0

Step 2: Graph the constraints.

Step 3: Find the feasible region.

Step 4: Find the coordinates of the vertices of the feasible region.

Vertices: (0, 0), (0, 5), (5, 0), (2.5, 2.5)

Step 5: Substitute the coordinates of the vertices into the objective function to determine the maximum and minimum values of Z.

Maximum Z: Z = 3(2.5) + 2(2.5) = 12.5

Minimum Z: Z = 3(0) + 2(0) = 0

Question:

Maximize Z=3x+4y subject to the constraints: x+y≤4,x≥0,y≥0

Answer:

Step 1: Write the objective function in standard form: Maximize Z = 3x + 4y

Step 2: Write the constraints in standard form: x + y ≤ 4 x ≥ 0 y ≥ 0

Step 3: Draw a graph of the constraints.

Step 4: Find the feasible region and identify the corner points.

The feasible region is the area bounded by the lines x + y = 4, x = 0, and y = 0. The corner points are (0,0), (0,4), and (4,0).

Step 5: Evaluate the objective function at each corner point and identify the maximum value.

At (0,0): Z = 0 At (0,4): Z = 16 At (4,0): Z = 12

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