Skip to main content

Graphical max/min min:Z=5x₁+3x₂

Short answer

Plot the lines 3x1+5x2=15 and 5x1+2x2=10 to find the feasible region. Corner points are (0,0), (2,0), (20/19, 45/19), (0,3). Evaluate Z=5x1+3x2: maximum Z=235/19 at (20/19, 45/19); minimum Z=0 at (0,0).

Graphically maximize and minimize Z = 5x₁ + 3x₂ subject to the constraints 3x₁ + 5x₂ ≤ 15, 5x₁ + 2x₂ ≤ 10, 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 + 5x₂ ≤ 153x₁ + 5x₂ = 15(5, 0)(0, 3)towards the origin
(2) 5x₁ + 2x₂ ≤ 105x₁ + 2x₂ = 10(2, 0)(0, 5)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, ….

   4|                                            
    |                                            
    |                                            
    |                                            
    |D                                           
    |......                                      
    |............                                
    |...............C                            
    |..................                          
    |...................                         
    |.....................                       
    |......................                      
    |.......................                     
    |.........................                   
    |..........................                  
    |............................                
   0|A............................B              
    +--------------------------------------------
     0                                          3

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 = 5x₁ + 3x₂
A(0, 0)x₁ = 0, x₂ = 00
B(2, 0)(2), x₂ = 010
C(20/19, 45/19)(1), (2)235/19
D(0, 3)(1), x₁ = 09

Step 4: Conclusion. The largest value is 235/19 at C(20/19, 45/19) and the smallest value is 0 at A(0, 0).

Answer: Maximum Z = 235/19 at x₁ = 20/19, x₂ = 45/19; Minimum Z = 0 at x₁ = 0, x₂ = 0.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Mathematics III chapter 5 questions →