已知实数 x,y 满足 x(x+y)=1+2y^2 ，求 5x^2-4y^2 的最小值

wintex

請問這個問題，謝謝。

xfhaoym

如果要用条件极值能解,有点麻烦.要用初等数学也可能解出.

pro

见照片解答。

luyuanhong

楼上 pro 的解答很好！已收藏。下面是另一种解法：

Nicolas2050

Initial Objective: 5.0
Final Objective: 3.9999653330068012
Solution
x1 = -0.999994960062743
x2 = -0.4999960669214395

Nicolas2050

# -*- coding: utf-8 -*-
"""
Created on Sun Jan  5 14:16:13 2020

Problem: Nonlinear Optimization¶
Solve the following nonlinear optimization problem:
   
已知实数 x,y 满足 x(x+y)=1+2y^2 ，求 5x^2-4y^2 的最小值
⑴ min: 5x^2-4y^2  
s.t.
-10≤x,y≤10
x(x+y)=1+2y^2
with initial conditions:
x0=(-1,1)

@author: NICOLAS TU
"""
from math import sqrt
import numpy as np
from scipy.optimize import minimize

def objective(x):
    return  5*(x[0])**2-4*(x[1])**2
def constraint1(x):
    sum_eq =1.0-(x[0])**2-(x[0]*x[1])+2*(x[1])**2   
    return sum_eq

# initial guesses
n = 2
x0 = np.zeros(n)
x0[0] =-1
x0[1] =1



# show initial objective
print('Initial Objective: ' + str(objective(x0)))

# optimize
b = (-10,10)
bnds = (b,b)
con1 = {'type': 'eq', 'fun': constraint1}
cons = ([con1])
solution = minimize(objective,x0,method='SLSQP',\
                    bounds=bnds,constraints=cons)
x = solution.x

# show final objective
print('Final Objective: ' + str(objective(x)))

# print solution
print('Solution')
print('x1 = ' + str(x[0]))
print('x2 = ' + str(x[1]))

'''
Initial Objective: 5.0
Final Objective: 3.9999653330068012
Solution
x1 = -0.999994960062743
x2 = -0.4999960669214395
'''

Nicolas2050

Initial Objective: 1.0
Final Objective: 4.000000000001245
Solution
x1 = 1.0000000291193183
x2 = 0.5000000727979802

掬一捧月光

luyuanhong

楼上 掬一捧月光 的解答很好！已收藏。