3
14
,
3
4
are
and
2
1
2
1
=
=
∗
∗
∗
∗
x
x
x
x
The
maximum value of the objective function is
10
3
14
2
3
4
5
.
0
=
×
+
×
=
z
, which was by [5].
Now consider a non-linear programming
problem, which differs from the linear
programming problem only in that the
objective function:
(
)
(
)
.
4
20
5
.
3
10
2
2
2
1
−
+
−
=
x
x
z
(5)
Imagine that it is desired to minimize the
objective function. Observe that here we have
a separable objective function. The graphical
solution of this problem is given in Fig.2
Fig. 2 Optimal solution by graphical method
The region representing the feasible solution
is, of course, precisely the same as that for