Добавил:
Upload Опубликованный материал нарушает ваши авторские права? Сообщите нам.
Вуз: Предмет: Файл:
СБОРНИК ЗАДАЧ ПО МАТЕМАТИКЕ ДЛЯ ТРАНСПОРТНЫХ СП...doc
Скачиваний:
36
Добавлен:
21.11.2019
Размер:
5.31 Mб
Скачать

Вариант 17

  1. f(X) = 2x1 + 3x2 max

-2x1 + 2x2 5

x1 – 4x2 3

x1 – x2 5

xj 0, j = 1,2.

  1. f(X) = x1x2 + x3 min

x1 + 2x2 + x3 + x4 = 3

x1 -4x2 – x4 = -2

xj 0, j = 1,2,3,4.

  1. f(X) =2x13x2 + x4 min

x1 + 2x2 – x3 =2

x3 + x4 =6

xj 0, j = 1,2,3,4.

  1. f(X) =2x32x4 x5 min

x1 -2x2 + x4 + x5 = 4

x2 + 3x3 – x4 + x5 = 2

xj 0, j = 1,2,3,4,5.

5. f(X) = 4x1 + 6x2 max

2x1 + x2 16

x1 + x2 10

xj 0, j = 1,2.

6. f(X) = 4x1 + 4x2 + 5x3 + 3x4 min

x1 + 2x2 + x3 - x4 + x5 = 10

3x1 + x2 + 3x3 + 2x4 - 2x5 = 12

- x1 + x2 + x4 + x5 = 10

xj 0, j = 1,2,3,4,5.

7. f(X) = - 2x1 + x2 max

x1 - 2x2 1

x1 + 2x2 3

xj 0, j = 1,2.

8. f(X) = 2x1 - x2 + x3 min

x1 + x2 – 2x3 5

x1 + x2 + x3 7

x1 - x2 - x3 = 1

xj 0, j = 1,2.

  1. f(X) =2x13x2 + x4 min

x1 + 2x2 – x3 =2

x3 + x4 =6

xj 0, j = 1,2,3,4.

  1. f(X) =2x32x4 x5 min

x1 -2x2 + x4 + x5 = 4

x2 + 3x3 – x4 + x5 = 2

xj 0, j = 1,2,3,4,5.

Вариант 18

  1. f(X) = x1+ x2 + x3 max

2x1 + x2 + x3 = 3

x1 + x2 5

xj 0, j = 1,2,3.

  1. f(X) = x1 + x2 max

x1 + x3 =2

x2 - x3 + x4 =1

xj 0, j = 1,2,3,4.

  1. f(X) = -2x1 + 3x3 max

2x1 + x2 + x3 =4

-x1 + 3x2 – x3 = 5

xj 0, j = 1,2,3.

  1. f(X) = 2x1x2 + 3x3 + x4 min

2x1 + x2 – 3x4 =10

x1 + x3 + x4 = 7

-3x1 – 2x3 + x5 = 4

xj 0, j = 1,2,3,4,5.

5. f(X) = 2x1 + 3x2 max

-2x1 + x2 6

x1 - 4x2 2

x1 - x2 5

xj 0, j = 1,2.

6. f(X) = -3x1 + 6x2+ 5x3 -5x4 + 6x5 min

2x1 - x2 + x3 + 2x4 - 3x5 = 4

x1 + x2 + 3x3 - x4 + 2x5 = 15

3 x1 - 2x2 + x3 + x4 + x5 = 4

xj 0, j = 1,2,3,4,5.

7. f(X) = x1 + 7x2 max

2x1 + 8x2 5

x1 – x2 1

xj 0, j = 1,2.

8. f(X) = x1 + 2x2 - x3 + x4 max

x1 + x2 – x4 1

x1 + x2 + x3 2

x1 + x3 - x4 =5

xj 0, j = 1,2,3,4.

  1. f(X) = -2x1 + 3x3 max

2x1 + x2 + x3 =4

-x1 + 3x2 – x3 = 5

xj 0, j = 1,2,3.

  1. f(X) = 2x1x2 + 3x3 + x4 min

2x1 + x2 – 3x4 =10

x1 + x3 + x4 = 7

-3x1 – 2x3 + x5 = 4

xj 0, j = 1,2,3,4,5.

Вариант 19

  1. f(X) = x1 + 2x2 max

x1 + x2 6

2x1 + 3x2 2

xj 0, j = 1,2.

  1. f(X) = x2x1 max

x1 + 3x2 12

3x1 – x2 6

3x1 + 4x2 0

xj 0, j = 1,2.

  1. f(X) = x1 + x2 + x3 max

x1 + x3 =2

2x1 + x2 =1

xj 0, j = 1,2,3.

  1. f(X) = 2x1 - x2 min

2x1 – x2 12

x2 + x1 6

xj 0, j = 1,2.

  1. f(X) = x1 - x2 min

x1 + 2x2 =3

x1 - 4x2 = -2

xj 0, j = 1,2.

  1. f(X) = 6x1 + 2x2+ 2x3 -5x4 - 3x5 min

2x1 + x2 + x3 - x4 = 14

x1 + x2 - x3 + x4 + x5 = 10

-4x1 - 2x2 + x3 + 3x4 + 2x5 = 6

xj 0, j = 1,2,3,4,5.

  1. f(X) = - x1 + 2x2 max

2x2 + x1 3

x1 + x2 1

xj 0, j = 1,2.

  1. f(X) = x1 + 2x2 + x3 min

x1 + x3 1

x1 + x2 + 2x3 5

x1 + x2 – x3 =2

xj 0, j = 1,2,3.

  1. f(X) = x1 + x2 + x3 max

x1 + x3 =2

2x1 + x2 =1

xj 0, j = 1,2,3.

  1. f(X) = 2x1 - x2 min

2x1 – x2 12

x2 + x1 6

xj 0, j = 1,2.