Mathematica從5.0開始就可以解複雜的規劃模型，主要用Maximize和Minimize這兩個函數。下面給出幾個例子，說明如何用這兩個函數來求解規劃問題。
- 線性規劃
( * linear programming * )
Maximize[{x1 + x2 + x3 + x4, ( * objective function * )
x5 + x6 + x7 + x8 >= 250000 , ( * conditions * )
x1 + x5 <= 380000 ,
x2 + x6 <= 265200 ,
x3 + x7 <= 408100 ,
x4 + x8 <= 130100 ,
2.85 * x1 - 1.42 * x2 + 4.27 * x3 - 18.49 * x4 >= 0 ,
2.85 * x5 - 1.42 * x6 + 4.27 * x7 - 18.49 * x8 >= 0 ,
16.5 * x1 + 2 * x2 - 4 * x3 + 17 * x4 >= 0 ,
7.5 * x5 - 7 * x6 - 13.0 * x7 + 8 * x8 >= 0 ,
x1 >= 0 , x2 >= 0 , x3 >= 0 , x4 >= 0 , x5 >= 0 , x6 >= 0 , x7 >= 0 , x8 >= 0 },
{x1, x2, x3, x4, x5, x6, x7, x8} ( * vars * )
]
計算結果：
{ 933400 ., {x1 -> 264938 ., x2 -> 135702 ., x3 -> 408100 ., x4 ->
124660 ., x5 -> 115062 ., x6 -> 129498 ., x7 -> 0 ., x8 -> 5440.01 }}
其中，933400為目标函數的最優值，緊接的表為目标函數達到最優值時的結果。
- 二次規劃
- 0-1規劃
( * 0 - 1 Programming * )
Maximize[{ 3 x1 - 2 x2 + 5 x3,
x1 + 2 x2 - x3 ≤ 2 ,
x1 + 4 x2 + x3 ≤ 4 ,
x1 + x2 ≤ 3 ,
4 x2 + x3 ≤ 6 ,
x1 == 1 || x1 == 0 ,
x2 == 1 || x2 == 0 ,
x3 == 1 || x3 == 0
},
{x1, x2, x3}
]
- 整數規劃
( * Integer Programming * )
Maximize[{ 40 x1 - 90 x2,
9 x1 + 7 x2 ≤ 56 ,
7 x1 + 20 x2 ≤ 70 ,
x1 ≥ 0 ,
x2 ≥ 0 ,
x1 ∈ Integers,
x2 ∈ Integers
}, {x1, x2}
]
- 求近似解（用NMaximize/NMinimize）
NMinimize[{ 8 x1 + 10 x2 + 7 x3 + 6 x4 + 11 x5 + 9 x6,
12 x1 + 9 x2 + 25 x3 + 20 x4 + 17 x5 + 13 x6 ≥ 60 ,
35 x1 + 42 x2 + 18 x3 + 31 x4 + 56 x5 + 49 x6 ≥ 150 ,
37 x1 + 53 x2 + 28 x3 + 24 x4 + 29 x5 + 20 x6 ≥ 125 ,
0 ≤ x1 ≤ 1 ,
0 ≤ x2 ≤ 1 ,
0 ≤ x3 ≤ 1 ,
0 ≤ x4 ≤ 1 ,
0 ≤ x5 ≤ 1 ,
0 ≤ x6 ≤ 1
}, {x1, x2, x3, x4, x5, x6}]
- 非線性規劃
- 分數規劃
- 混合規劃
Maximize[{ 120 x1 + 10 x2 + 100 x3 - 5000 y1 - 2000 y2 - 1000 y3,
5 x1 + x2 + 4x3 ≤ 2000 ,
3 x1 ≤ 300 y1,
0.5 x2 ≤ 300 y2,
2 x3 ≤ 300 y3,
x1 ≥ 0 , x2 ≥ 0 , x3 ≥ 0 ,
x1 ∈ Integers, x2 ∈ Integers, x3 ∈ Integers,
y1 == 0 || y1 == 1 ,
y2 == 0 || y2 == 1 ,
y3 == 0 || y3 == 1
}, {x1, x2, x3, y1, y2, y3}
]
- 一個特殊的例子（幫助中說的原本的用法）