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Unformatted text preview: When y=1, f(x, 1) = which has local minimum at (+√3/2). When y=1, f(x, 1) = for all x. 2/3 EE5239 HW1 So the global minima are f(+√3/2,1) = . 1.1.3 (a) . for (b) , so (0, 0) is a stationary point. i. Consider the line y=0, f(0, z) = f(0, 0) = 0 for , (0,0) is a local minimum along line y=0. ii. Consider the line z =0, f(y, 0) = f(0, 0) = 0 for , (0,0) is a local minimum along line z=0. iii. Now consider the general case: along line , . + We have = 2 > 0, so (0,0) is a local minimum along line , Then we can say that (0,0) is a local minimum of f along every line that pass through (0,0). If p < m <q, then , so (0,0) is not a local minimum of f. 1.1.4 Let 2, then x=1. for all x > 0, which means that is a convex function for . 3/3 EE5239 HW1 Then the global minimum is reached at , x=1. f(x) f(1) = 0 for all x > 0 . 4/3...
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 Spring '10
 DR.RAHULA
 Critical Point, Optimization, Fermat's theorem

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