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===Solution=== | ===Solution=== | ||
Let <math>t_1=x_1-2</math>, <math>t_2=x_2+1</math><br> | Let <math>t_1=x_1-2</math>, <math>t_2=x_2+1</math><br> | ||
| − | so that <math>g(t_1,t_2)=\dfrac{1}{t_1^2+t_2^2+3}|t_1=0,t_2</math> would have some convex property with <math>f(x_1,x_2)=\dfrac{1}{(x_1-2)^2+(x_2+1)^2+3}|x_1=2,x_1=-1</math><br> | + | so that <math>g(t_1,t_2)=\dfrac{1}{t_1^2+t_2^2+3}|t_1=0,t_2=0</math> would have some convex property<br> |
| + | with <math>f(x_1,x_2)=\dfrac{1}{(x_1-2)^2+(x_2+1)^2+3}|x_1=2,x_1=-1</math><br> | ||
---- | ---- | ||
[[QE2016_AC-3_ECE580|Back to QE AC question 3, August 2016]] | [[QE2016_AC-3_ECE580|Back to QE AC question 3, August 2016]] | ||
[[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]] | [[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]] | ||
Revision as of 21:00, 18 February 2019
Automatic Control (AC)
Question 3: Optimization
August 2016 Problem 3
Solution
Let $ t_1=x_1-2 $, $ t_2=x_2+1 $
so that $ g(t_1,t_2)=\dfrac{1}{t_1^2+t_2^2+3}|t_1=0,t_2=0 $ would have some convex property
with $ f(x_1,x_2)=\dfrac{1}{(x_1-2)^2+(x_2+1)^2+3}|x_1=2,x_1=-1 $
