Методы оптимизации / Задания для самостоятельной работы 4 по курсу Методы оптимизации

f(Х)= 3x₁³ – x₁ – х₂³ – 3х₂² - 1 max ; Х⁰ = (0, 0)

f(Х)= x₁² – 4x₁ + 2х₂² + 2х₂ min ; Х⁰ = (1, 0)

f(Х)= x₁² x₂ – 2x₁x₂ + x₁/х₂ min ; Х⁰ = (1, 2)

f(Х)= 9x₁² – 90x₁ + 16х₂² - 128х₂ min ; Х⁰ = (1, 0)

f(Х)= 2x₁² – 12x₁ + х₂² min ; Х⁰ = (5, 3)

f(Х)= 2x₁² + x₂ ² – 3x₁x₂ min ; Х⁰ = (2, 2)

f(Х)= (x₁ – 2)² – (х₂ – 3)² max ; Х⁰ = (6, 4)

f(Х)= 4x₁ – 8x₂ - 2x₁² - 2х₂² max ; Х⁰ = (5, 10)

f(Х)= 8x₁ + 32x₂ - 2x₁² - 4х₂² max ; Х⁰ = (6, 6)

f(Х)= 2x₁ - x₁² - х₂² max ; Х⁰ = (3, 2)

f(Х)= 4x₁ + 2x₂ - x₁² - х₂² + 5 max ; Х⁰ = (4, 5)

f(Х)= 10x₁ – 16x₂ - x₁² - х₂² max ; Х⁰ = (1, 2)

f(Х)= 6x₁ - 2x₁² + 2x₁  x₂ - 2х₂² max ; Х⁰ = (0, 0)

f(Х)= 6x₁ + 4x₂ + 2х₃ - 3x₁² - 2x₁² –(1/3)  х₃² max ; Х⁰ = (0, 0, 3)

f(Х)= x₁² + 4х₂² - 1 min ; Х⁰ = (1, 1)

f(Х)= 3x₁³ – x₁ + х₂³ - 3х₂² - 1 min ; Х⁰ = (0, 0)

f(Х)= x₁²  x₂ - 2x₁  x₂ + x₁/х₂ min ; Х⁰ = (1, 2)

f(Х)= 9x₁² – 90x₁ + 16х₂² - 128х₂ min ; Х⁰ = (0, 0)

f(Х)= 2x₁² + х₂² - 12х₁ min ; Х⁰ = (5, 3)

f(Х)= x₁ - x₂ + 5х₃ - x₁² + 2x₂² + (1/2)  х₃² max ; Х⁰ = (0, 1, 3)

f(Х)= 2x₁² + х₂² - 3х₁ х₂ min ; Х⁰ = (2, 2)

f(Х)= (x₁ - 2)² + (х₂ + 3)² max ; Х⁰ = (6, 4)

f(Х)= 4x₁ + 8x₂ - 2x₁² - 2х₂² max ; Х⁰ = (5, 10)

f(Х)= 8x₁ - 32x₂ - 2x₁² + 4х₂² max ; Х⁰ = (6, 6)

f(Х)= 2x₁ - x₁² + 2х₂² max ; Х⁰ = (3, 2)

f(Х)= 4x₁ - 2x₂ - x₁² - х₂² + 1 max ; Х⁰ = (4, 5)

f(Х)= 10x₁ – 16x₂ - x₁² + х₂² max ; Х⁰ = (1, 2)