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[[Category:optimization]]
 
[[Category:optimization]]
  
=QE2013_AC-3_ECE580-4=
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=QE2013_AC-3_ECE580-5=
  
 
:[[QE2013_AC-3_ECE580-1|Part 1]],[[QE2013_AC-3_ECE580-2|2]],[[QE2013_AC-3_ECE580-3|3]],[[QE2013_AC-3_ECE580-4|4]],[[QE2013_AC-3_ECE580-5|5]]
 
:[[QE2013_AC-3_ECE580-1|Part 1]],[[QE2013_AC-3_ECE580-2|2]],[[QE2013_AC-3_ECE580-3|3]],[[QE2013_AC-3_ECE580-4|4]],[[QE2013_AC-3_ECE580-5|5]]
  
 
<br> '''Solution: ''' <br>
 
<br> '''Solution: ''' <br>
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 +
From the constraint, it can be seen that:
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<math>x_1 = x_3 = -x_2 </math>
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Substitute into the objective function:
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<math>f(x) = x_2 (x_1 + x_3) = -2 x_2^2 </math>
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Therefore it has a maximizer but no minimizer (f(x) goes to <math>-\infty</math> as <math>|x_2|</math> increases)
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The maximizer is <math>x_1 = x_2 = x_3 = 0</math>.  There f(x) reaches the maximum value of 0.
  
  
 
[[ QE2013 AC-3 ECE580|Back to QE2013 AC-3 ECE580]]
 
[[ QE2013 AC-3 ECE580|Back to QE2013 AC-3 ECE580]]

Revision as of 10:08, 27 January 2015


QE2013_AC-3_ECE580-5

Part 1,2,3,4,5


Solution:

From the constraint, it can be seen that:

$ x_1 = x_3 = -x_2 $

Substitute into the objective function:

$ f(x) = x_2 (x_1 + x_3) = -2 x_2^2 $

Therefore it has a maximizer but no minimizer (f(x) goes to $ -\infty $ as $ |x_2| $ increases)

The maximizer is $ x_1 = x_2 = x_3 = 0 $. There f(x) reaches the maximum value of 0.


Back to QE2013 AC-3 ECE580

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva