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Graphical method max:z=2x₁+3x₂

Short answer

The feasible region is unbounded. Along the edge x₁ − x₂ = 2, Z = 5t − 6 grows without limit as t → ∞, so Z has no finite maximum.

Maximize z = 2x₁ + 3x₂ subject to constraints x₁ − x₂ ≤ 2, x₁ + x₂ ≥ 4 and x₁, x₂ ≥ 0 graphically.

Step 1: Draw the constraint lines. Replace each inequality by an equation, find the intercepts and decide which side of the line is allowed (test the origin).

ConstraintLineMeets x-axisMeets y-axisAllowed side
(1) x₁ − x₂ ≤ 2x₁ − x₂ = 2(2, 0)(0, −2)towards the origin
(2) x₁ + x₂ ≥ 4x₁ + x₂ = 4(4, 0)(0, 4)away from the origin

Also x₁ ≥ 0 and x₂ ≥ 0, so the region lies in the first quadrant.

Step 2: Feasible region. The region satisfying all constraints is the shaded polygon below (dots), with corner points marked A, B, C, ….

   9|............................................
    |............................................
    |............................................
    |............................................
    |..........................................  
    |........................................    
    |.....................................       
    |..................................          
    |................................            
    |B............................               
    |   .......................                  
    |      .................                     
    |         ............                       
    |            ......                          
    |              A                             
    |                                            
   0|                                            
    +--------------------------------------------
     0                                          9

Step 3: Corner points and the objective value. The optimum of a linear function over a convex polygon occurs at a corner point.

PointCoordinatesLines meeting thereZ = 2x₁ + 3x₂
A(3, 1)(1), (2)9
B(0, 4)(2), x₁ = 012

Step 4: Is the region bounded? The corner points are A(3, 1) and B(0, 4) (the origin and the axis points are cut off by x₁ + x₂ ≥ 4 and x₁ − x₂ ≤ 2). Beyond them the feasible region extends to infinity: no constraint stops x₁ and x₂ from increasing together. For instance, along the line x₁ − x₂ = 2 the points (t, t − 2), t ≥ 3, are all feasible, and there

Z = 2t + 3(t − 2) = 5t − 6,

which grows without limit as t → ∞. Moving the line Z = constant further from the origin never leaves the region.

Answer: The objective function is unbounded above: Z has no finite maximum. (The corner values are Z = 9 at A(3, 1) and Z = 12 at B(0, 4), but Z can be increased indefinitely along the unbounded edge.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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