Question

1- solve the following LPP :

Minimize Z = 12x + 3y

Subject to 4x + 6y > 24,000

X + y > 5,000

8x + 2y > 16,000

And x> 0, y> 0

Answer 1

I've used Graphical method. If you need any other method just comment below I will change it straight away.
PLEASE CONSIDER LIKING THIS ANSWER. THANK YOU!

Solution:
Problem is

MIN Zx = 12 x1 + 3 x2

subject to

4 x1 + 6 x2 ≥ 24000 x1 + x2 ≥ 5000 8 x1 + 2 x2 ≥ 16000

and x1,x2≥0;

Hint to draw constraints

1. To draw constraint 4x1+6x2≥24000→(1)

Treat it as 4x1+6x2=24000

When x1=0 then x2=?
⇒4(0)+6x2=24000
⇒6x2=24000
⇒x2=240006=4000

When x2=0 then x1=?
⇒4x1+6(0)=24000
⇒4x1=24000
⇒x1=240004=6000

x1	0	6000
x2	4000	0

2. To draw constraint x1+x2≥5000→(2)

Treat it as x1+x2=5000

When x1=0 then x2=?
⇒(0)+x2=5000
⇒x2=5000

When x2=0 then x1=?
⇒x1+(0)=5000
⇒x1=5000

x1	0	5000
x2	5000	0

3. To draw constraint 8x1+2x2≥16000→(3)

Treat it as 8x1+2x2=16000

When x1=0 then x2=?
⇒8(0)+2x2=16000
⇒2x2=16000
⇒x2=160002=8000

When x2=0 then x1=?
⇒8x1+2(0)=16000
⇒8x1=16000
⇒x1=160008=2000

x1	0	2000
x2	8000	0

The value of the objective function at each of these extreme points is as follows:

Extreme Point Coordinates (x1,x2)	Lines through Extreme Point	Objective function value Z=12x1+3x2
A(0,8000)	3→8x1+2x2≥16000 4→x1≥0	12(0)+3(8000)=24000
B(1000,4000)	2→x1+x2≥5000 3→8x1+2x2≥16000	12(1000)+3(4000)=24000
C(3000,2000)	1→4x1+6x2≥24000 2→x1+x2≥5000	12(3000)+3(2000)=42000
D(6000,0)	1→4x1+6x2≥24000 5→x2≥0	12(6000)+3(0)=72000

The minimum value of the objective function Z = 24000 occurs at 2 extreme points.
Hence, problem has multiple optimal solutions and min Z = 24000.

*****Please please please LIKE THIS ANSWER, so that I can get a small benefit, Please*****