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Linear programming


              Example 3.5
              Solve graphically the following linear programming problem.
              Minimize f (x, y) = 2x + 5y
              Subject to:  2x +  4y ≤  8
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                         3x +  2y ≤  6
                         x +≥   1
                             y
                         x ≥  0, y ≥  0
              Solution
              The graph of the given constraints is as follows.

                                           y
                                               x=0
                                            3


                                            2  C(0, 2)
                                                    D 1, 3
                                     x+y=1
                                                        2
                                            1
                                       B(0, 1)  Feasible region  2x+4y=8
                                             A(1, 0)       E(2, 0)       y=0
                              -2     -1     0 0    1      2      3      4   x


                                          -1                3x+2y=6




                The values of the objective function at each of the extreme points are shown
                in the following table.

                          Corner points of the       Values of the objective function
                            feasible region        f (, )xy = 2x + 5y
                          A(1, 0)                  f = 2(1) + 5(0) = 2

                          B(0, 1)                  f = 2(0) + 5(1) = 5                              Mathematics for Secondary Schools
                          C(0, 2)                  f = 2(0) + 5(2) =10
                                                             3
                           ⎛   3 ⎞                               1
                                                       ( ) 5+
                          D 1,   ⎟                 f =  21    =  9
                           ⎜
                                                             2
                           ⎝   2 ⎠                               2
                          E(2, 0)                  f = 2(2) + 5(0) = 4
               From the table, the objective function has the smallest value at point A (1, 0).
               Therefore, the minimum value is 2.




                 Student\s Book Form Three          67



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     MATHEMATIC F3 SB.indd   67
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