Can you please help in solving the linear equation graphically with clear steps Special case 2:Alternative optimal solutions (1)Max Z= 6x, + 8x₂s.t.2X₁ X₂ ≤ 503x₂ +4x₂ ≤ 180- X₁ +2X₂ ≤ 40X₁ X₂ ≥ 0-There are two basic optimalsolutions: x* = (20, 30)***= (34 6/11, 19 1/11)x₂80706050403020and any point on the linesegment bounded by x* and x** 10is an alternative optimalsolution.For any of these points, themaximum is: Z* = 360Feasible solution Region+*2x₁-x₂=50***-x₂+2x₂=403x,+4x₂=180(6,8)10 20 30 40 50 60 70ISO-profit line80X₁ Special case 3:Alternative optimal solutions (2)Max Z=- 2x₁ + x₂s.t.X₁ - 2X₂ ≤ 4-2X₁ + X₂ ≤ 3X₁, X2₂ ≥ 08764There is a unique basicoptimal solution: x* = (0, 3)and any point on the half-lineor ray bounded by x* is analternative optimal solution. 37For any of these points, the2maximum is: Z* = 3(-2,1)1-2x₁+x₂=3ISO-profit lineFeasible Solution Region1 2 3 4 5 6 7x₁- 2x₂=48 9 10X₂

Answered: Max Z= 6x, + 8x₂ s.t. 2X₁ X₂ ≤ 50 3x₂…

Practical Management Science

6th Edition

ISBN:9781337406659

Author:WINSTON, Wayne L.

Publisher:WINSTON, Wayne L.

Chapter2: Introduction To Spreadsheet Modeling

Section: Chapter Questions

Problem 20P: Julie James is opening a lemonade stand. She believes the fixed cost per week of running the stand...

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Can you please help in solving the linear equation graphically with clear steps

Special case 2:
Alternative optimal solutions (1)
Max Z= 6x, + 8x₂
s.t.
2X₁ X₂ ≤ 50
3x₂ +4x₂ ≤ 180
- X₁ +2X₂ ≤ 40
X₁ X₂ ≥ 0
-
There are two basic optimal
solutions: x* = (20, 30)
***= (34 6/11, 19 1/11)
x₂
80
70
60
50
40
30
20
and any point on the line
segment bounded by x* and x** 10
is an alternative optimal
solution.
For any of these points, the
maximum is: Z* = 360
Feasible solution Region
+*
2x₁-x₂=50
***
-x₂+2x₂=40
3x,+4x₂=180
(6,8)
10 20 30 40 50 60 70
ISO-profit line
80
X₁

Special case 3:
Alternative optimal solutions (2)
Max Z=- 2x₁ + x₂
s.t.
X₁ - 2X₂ ≤ 4
-2X₁ + X₂ ≤ 3
X₁, X2₂ ≥ 0
8
7
6
4
There is a unique basic
optimal solution: x* = (0, 3)
and any point on the half-line
or ray bounded by x* is an
alternative optimal solution. 37
For any of these points, the
2
maximum is: Z* = 3
(-2,1)
1
-2x₁+x₂=3
ISO-profit line
Feasible Solution Region
1 2 3 4 5 6 7
x₁- 2x₂=4
8 9 10
X₂

Expert Solution

Step 1 Introduction:-

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Conceptual Introduction:-

Linear programming is a mathematical technique that is also used in operations management departments. This technique is commonly used to find the best outcome with the maximum profit. Linear programming has various methods; each method is chosen by the companies as per their requirements or based on different factors.

Steps to solve the graphical problem:-

Equalize each constraint and determine the values of x1 and x2.
Plot all these values on the graph.
Find the corner points along the feasible region.
Determine the minimum/maximum value from the outcomes.

Below is the solution by using these steps.

Plotting values on the graph:-

Plotting all the above-determined values of each constraint on the graph. Finding the feasible region. The constrain line that is (<=) will be shared towards the origin (0,0) and the constraint that has (>=) will shade opposite to the origin (0,0). The common region of all their three lines of constraint will be the feasible region.

Corner points of the feasible region:- Fill the values of x1 and x2 in the objective function. Then choose the minimum/maximum (as per the objective function) value from the outcomes of the objective function.