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Question:
1. Solve the following problem using the simplex method.
Maximise p = 4 x + 6 y + 3 z
Subjected to the following constraints:
(N+1) x + 4 y + 3 z ≤ 12
4 x + 3 y + 2 z ≤ 10
2 x + 4 y + z ≤ 8
x≥ 0, y≥ 0 and z≥ 0
2. By using the Lagrange multiplier method, find the dimensions of a rectangular box of volume V = 1000 × (N+1)2 cm3 for which the total length of the 12 edges is minimum.
3. Consider the function f(x) = x4 – 14x3 + (N+51) x2 - 70x
Use the golden section search technique to find the value of x that minimises f over the range 0 ≤ x ≤ 2. Locate the x value within an interval of 15% uncertainty of the initial search interval. Present your answer in a tabular format.
4. A manufacturing company produces two types of products, Product A and Product B. The various restrictions, requirements and unit profits are shown below. By using the analytical linear programming method determine how many of Product A and Product B should be produced to maximise the profit. Present your answer in a tabular format.
† Individual assignment
Profit Raw Material Electricity Labour
Product A 3 5 1 1
Product B 4 2 1 2
Amount available ---- 31 + N 10
Answer:

As given in question,

Maximize p = 4x + 6y + 3z subject to

6x + 4y + 3z <= 12

4x + 3y +2z <= 10

2w + 4y + z <= 8

x>=0

y>=0

z>=0

Suppose s­­1, s2, s3, s4, s5 and s6  are the slack variable, the above equation can be tabulated as follows

Table

#1

 

 

 

 

 

 

 

 

 

x

y

z

s1

s2

s3

s4

s5

s6

p

 

6

4

3

1

0

0

0

0

0

0

12

4

3

2

0

1

0

0

0

0

0

10

2

4

1

0

0

1

0

0

0

0

8

4

6

3

0

0

0

0

0

0

0

 

 

Further row and column operation can be done as follows

Tableau

#1

                 

x

y

z

s1

s2

s3

s4

s5

s6

p

 

6

4

3

1

0

0

0

0

0

0

12

4

3

2

0

1

0

0

0

0

0

10

0

4

1

0

0

1

0

0

0

0

8

1

0

0

0

0

0

-1

0

0

0

0

0

1

0

0

0

0

0

-1

0

0

0

0

0

1

0

0

0

0

0

-1

0

0

-4

-6

-3

0

0

0

0

0

0

1

0

 

Tableau

#2

                 

x

y

z

s1

s2

s3

s4

s5

s6

p

 

6

4

3

1

0

0

0

0

0

0

12

4

3

2

0

1

0

0

0

0

0

10

0

4

1

0

0

1

0

0

0

0

8

-1

0

0

0

0

0

1

0

0

0

0

0

1

0

0

0

0

0

-1

0

0

0

0

0

1

0

0

0

0

0

-1

0

0

-4

-6

-3

0

0

0

0

0

0

1

0

 

Tableau

#3

                 

x

y

z

s1

s2

s3

s4

s5

s6

p

 

6

4

3

1

0

0

0

0

0

0

12

4

3

2

0

1

0

0

0

0

0

10

0

4

1

0

0

1

0

0

0

0

8

-1

0

0

0

0

0

1

0

0

0

0

0

-1

0

0

0

0

0

1

0

0

0

0

0

1

0

0

0

0

0

-1

0

0

-4

-6

-3

0

0

0

0

0

0

1

0

 

Tableau

#4

                 

x

y

z

s1

s2

s3

s4

s5

s6

p

 

6

4

3

1

0

0

0

0

0

0

12

4

3

2

0

1

0

0

0

0

0

10

0

4

1

0

0

1

0

0

0

0

8

-1

0

0

0

0

0

1

0

0

0

0

0

-1

0

0

0

0

0

1

0

0

0

0

0

-1

0

0

0

0

0

1

0

0

-4

-6

-3

0

0

0

0

0

0

1

0

 

Tableau

#5

                 

x

y

z

s1

s2

s3

s4

s5

s6

p

 

6

0

2

1

0

-1

0

0

0

0

4

4

0

1.3

0

1

-0.75

0

0

0

0

4

0

1

0.25

0

0

0.25

0

0

0

0

2

-1

0

0

0

0

0

1

0

0

0

0

0

0

0.25

0

0

0.25

0

1

0

0

2

0

0

-1

0

0

0

0

0

1

0

0

-4

0

-1.5

0

0

1.5

0

0

0

1

12

 

Tableau

#6

                 

x

y

z

s1

s2

s3

s4

s5

s6

p

 

1

0

0.33

0.17

0

-0.17

0

0

0

0

0.67

0

0

-0.083

-0.67

1

-0.083

0

0

0

0

1.3

0

1

0.25

0

0

0.25

0

0

0

0

2

0

0

0.33

0.17

0

-0.17

1

0

0

0

0.67

0

0

0.25

0

0

0.25

0

1

0

0

2

0

0

-1

0

0

0

0

0

1

0

0

0

0

-0.17

0.67

0

0.83

0

0

0

1

15

 

Table

#7

                 

x

y

z

s1

s2

s3

s4

s5

s6

p

 

3

0

1

0.5

0

-0.5

0

0

0

0

2

0.25

0

0

-0.63

1

-0.12

0

0

0

0

1.5

-0.75

1

0

-0.13

0

0.38

0

0

0

0

1.5

-1

0

0

0

0

0

1

0

0

0

0

-0.75

0

0

-0.13

0

0.38

0

1

0

0

1.5

3

0

0

0.5

0

-0.5

0

0

1

0

2

0.5

0

0

0.75

0

0.75

0

0

0

1

15

Since, we can see that, no positive unit solution for x variable, in this condition it can be taken as 0, y = 1.5 and z = 2, the maximum value of P whiting the given condition is 15

Solution 2

As given in problem,

We have to minimize total length i.e. f(x,y,z) = 4x+4y+4z, in which x, y and z are length width and height

Which is subjected to the constant g(x,y,z) = xyz = 1000 (5+1)2 = 36000 cm3,

By implementing Lagrange multiplier, we get

                        f(x­0,y0,z0) = λ f (x­0,y0,z0)

Using this equation in the above minimization problem

            4x + 4y =  λxy             ….. (i)

            4y + 4z =  λyz             ….. (ii)

            4x + 4z =  λzx             ….. (iii)

Since the constraints x  0, y  0, and z  0, in this condition, we have to solve λ for each equation in pairs we get,

                             ….. (iv)

                              ….. (v)

                  ….. (vi)

From equation (iv) and (v) we get,

              =   or x =z

            Similarly, equation (v) and (vi) we get

                        X =z, i.e. x =y=z

From the above solution it’s clear that the sum will be minimum when all dimension will be equal, i.e. the box will be in the form of cuboid

                        Putting g the value in constrain

                                    Xyz = x3 = 36000

                        Or,   x = , therefore all side will be x = =            Ans

Solution 3

As given in question,

The value of x range as given =  = [0, 2]

The steps to be followed is as given

1         

Now applying the gold section method at  (Suppose)

2  Now putting the value in function,        

           We have to minimise the problem          

            F(0) = 4

Now f(0) = 0, f(0.7639) = -24.361,  f(1.2631) = -18.958

            Value less than f(2) = 4

            X3 0.4722 = (0+L3*)           

            F(0) = 0 à  f(x1) = -24.361           

                        F(0) = 0    

Function

L1

2 (0, 1.2361)

L3 (0.4722, 1.2361)

f(x1)

(0,2)

-24.361

 

f(x2)

(0,2)

-18.958

 

f(x3)

(0,2)

 

-21.1

Solution 4
As given in question,

The constraints can be tabulated as follows

 

Profit

Raw Material

Electricity

Labor

Product A

3

5

1

1

Product B

4

2

1

2

Amount available

----

36

10

16

Suppose the number of product A =x

And No of product B = y

            Then, Profit (Z) = 3x +4y, we must maximize Z to solve the problem

            The constraints can be written as

            5x+2y  36       ……… (i)

            x+y  10            ……… (ii)

            x+2y  16           ……... (iii) 

Suppose s1 , s2, s3, s4 , s5  are the slack variable for the given above equation.

After performing different operation in row and column

Table

1

             

x

y

s1

s2

s3

s4

s5

p

 

5

2

1

0

0

0

0

0

36

1

1

0

1

0

0

0

0

10

1

2

0

0

1

0

0

0

16

1

0

0

0

0

-1

0

0

0

0

1

0

0

0

0

-1

0

0

-3

-4

0

0

0

0

0

1

0

 

Table

2

             

x

y

s1

s2

s3

s4

s5

p

 

5

2

1

0

0

0

0

0

36

1

1

0

1

0

0

0

0

10

1

2

0

0

1

0

0

0

16

-1

0

0

0

0

1

0

0

0

0

1

0

0

0

0

-1

0

0

-3

-4

0

0

0

0

0

1

0

 

Table

3

             

x

y

s1

s2

s3

s4

s5

p

 

5

2

1

0

0

0

0

0

36

1

1

0

1

0

0

0

0

10

1

2

0

0

1

0

0

0

16

-1

0

0

0

0

1

0

0

0

0

-1

0

0

0

0

1

0

0

-3

-4

0

0

0

0

0

1

0

 

Table

4

             

x

y

s1

s2

s3

s4

s5

p

 

4

0

1

0

-1

0

0

0

20

0.5

0

0

1

-0.5

0

0

0

2

0.5

1

0

0

0.5

0

0

0

8

-1

0

0

0

0

1

0

0

0

0.5

0

0

0

0.5

0

1

0

8

-1

0

0

0

2

0

0

1

32

 

Table

5

             

x

y

s1

s2

s3

s4

s5

p

 

0

0

1

-8

3

0

0

0

4

1

0

0

2

-1

0

0

0

4

0

1

0

-1

1

0

0

0

6

0

0

0

2

-1

1

0

0

4

0

0

0

-1

1

0

1

0

6

0

0

0

2

1

0

0

1

36

After making positive in all bottom line It was found that the value of x and are 4 and 6 respectively, the maximum profit that can be made is 36. This is tabulated below.           

 

Profit

Raw Material

Electricity

Labor

Product A

12

20

4

4

Product B

24

12

6

12

Amount available

----

32 <=36

10

16

References

Aczel Amir D.  Complete business statistics. NJ :: Morristown, , 2012.

Shuqin Yang. Applications of Excel in teaching Statistics. 2005.

Berenson Mark. 2 Basic Business Statistics. Sydney :: Pearson. 2012.

Broadman Bart.  DBS Group Holdings Ltd | Annual Report 2015. Singapore :: DBS Pub. 2014.

Barron's.  Business Statistics. New York :: 2010.

Krehbiel Heather Haskin & Timothy. Business statistics at the top 50 US business programmes. Dubline:: NSK. 2011.

Chng Pamela.  Bettr Barista Coffee Academy. Singapore : :Singa publisher. 2015.

Douglas Lind. William Marchal. Samuel Wathen. Basic Statistics for Business and Economics. New York : :McGraw-Hill Higher Education. 2012.

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My Assignment Help. (2020). Optimization Problems With Simplex And Lagrange Methods. Retrieved from https://myassignmenthelp.com/free-samples/mcen4011-engineering-design-methodology/dimensions-of-a-rectangular.html.

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[Accessed 24 July 2024].

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