Consider the following LPP.:

Max Z = 15x₁ + 10x₂

Subject to the constraints

4x₁ + 6x₂ ≤  360

3x1 + 0x2 ≤  180

0x1 + 5x2 ≤  200

x1, x2 ≥ 0

The solution of the LPP using Graphical solution-technique is :

The correct answer is option 2.

EXPLANATION 

4x₁ + 6x₂ ≤  360
if x₁=0=> 6x₂=360 => x2=60 (0,60)
if x₂=0=> 4x1=30  => x1=90 (90,0)
3x₁ + 0x2 ≤  180
3x1=180 => x1=60 =(60,0)
0x1 + 5x2 ≤  200
 5x2=200 =>x2=40= (0,40)

for B coordinates , consider 4x1 + 6x2 = 360

Put x1=60,

4x60+6x2=360

x2=20

At O point, Z = 15x1 + 10x2 , Z = 0×15 + 0×10, Z=0

At A point, Z = 15x1 + 10x2 , Z = 60×15 + 0×10, Z=900

At B point, Z = 15x1 + 10x2 , Z = 60×15 + 20×10, Z=1100

At C point, Z = 15x1 + 10x2 , Z = 30×15 + 40×10, Z=850

At D point, Z = 15x1 + 10x2 , Z = 0×15 + 40×10, Z=400

∴ Hence the correct answer is x1 = 60, x2​ = 20 and Z = 1100.