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Graphical method max:Z=7x₁+10x₂

Short answer

Plot lines 3x₁+x₂=9, x₁+x₂=8; feasible region corners: (0,0) Z=0, (3,0) Z=21, (0.5,7.5) Z=78.5, (0,8) Z=80. Maximum Z=80 at x₁=0, x₂=8.

Applying the graphical method, maximize Z = 7x₁ + 10x₂ subject to the constraints 3x₁ + x₂ ≤ 9, x₁ + x₂ ≤ 8, x₁, x₂ ≥ 0.

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) 3x₁ + x₂ ≤ 93x₁ + x₂ = 9(3, 0)(0, 9)towards the origin
(2) x₁ + x₂ ≤ 8x₁ + x₂ = 8(8, 0)(0, 8)towards 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|                                            
    |                                            
    |D.                                          
    |.....C.                                     
    |.........                                   
    |...........                                 
    |.............                               
    |...............                             
    |.................                           
    |...................                         
    |.....................                       
    |.......................                     
    |.........................                   
    |...........................                 
    |.............................               
    |...............................             
   0|A...............................B           
    +--------------------------------------------
     0                                          4

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 = 7x₁ + 10x₂
A(0, 0)x₁ = 0, x₂ = 00
B(3, 0)(1), x₂ = 021
C(1/2, 15/2)(1), (2)157/2
D(0, 8)(2), x₁ = 080

Step 4: Conclusion. The largest value of Z is 80, at D(0, 8).

Answer: Maximum Z = 80 at x₁ = 0, x₂ = 8.

✓ Answer checked by Claude Opus 5.5 (high effort)

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