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:[[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> | ||
+ | |||
+ | From the constraint, it can be seen that: | ||
+ | |||
+ | <math>x_1 = x_3 = -x_2 </math> | ||
+ | |||
+ | Substitute into the objective function: | ||
+ | |||
+ | <math>f(x) = x_2 (x_1 + x_3) = -2 x_2^2 </math> | ||
+ | |||
+ | Therefore it has a maximizer but no minimizer (f(x) goes to <math>-\infty</math> as <math>|x_2|</math> increases) | ||
+ | |||
+ | 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
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.