HW1_sol - When y=1, f(x, 1) = which has local minimum at...

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EE5239 HW1 1.1.1 = 0 , which is the stationary point for a given scalar . . But when has no solution. 1.1.2 (a) So , For (-2,0) and (2,0), we have f(x, y) = 0, and f(x, y) for (x,y) , so they are two global minima. For (0,0), has two eigenvalues -16 and 2, so it is neither a local maximum nor minimum. (b) , k = 0, , k = 0, So all stationary points are , k = 0, = When k = 0, … , > 0, ( ) is local minimum. When k = … , eigenvalues , 1/3
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EE5239 HW1 so ( ) = (0, ) is just a stationary point. (c) If x = -y, then there is no feasible solution. If x = y, then cosx + cos2x = 0 cosx = -1 or . = < 0 is local maximum. = f( ) 0 and f( ) for a small positive number . So is only a stationary point. = > 0 is a local minimum. (d) = is a stationary point, but it is neither a local maxima nor a local minimum. (e) Due to (d), the exterma of f(x,y) should be at the boundary y=-1 or y=1.
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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 &gt; 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 &lt; m &lt;q, then , so (0,0) is not a local minimum of f. 1.1.4 Let -2, then x=1. for all x &gt; 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 &gt; 0 . 4/3...
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HW1_sol - When y=1, f(x, 1) = which has local minimum at...

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