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THE UNIVERSITY OF NEW SOUTH WALESSCHOOL OF MATHEMATICS AND STATISTICSMATH3161/MATH5165 — OPTIMIZATION – Session 1, 2014Problem Sheet 4 Solutions – Unconstrained Problems1. Find all the stationary points of the following functions and identify, if possible, thelocal minimizers, local maximizers and saddle points.(a)f(x) =x21+x22+x23+x1x2+x1x3+x2x3-7x1-8x2-9x3+ 101AnswerThe gradient and Hessian offaref(x) =2x1+x2+x3-7x1+ 2x2+x3-8x1+x2+ 2x3-9,2f(x) =211121112The linear systemf(x) =0corresponds to the augmented matrix211712181129R1R3112912182117R2R2-R1R3R3-2R1112901-1-10-1-3-11R3R3+R1112901-1-100-4-12upon using row operations to reduce it to row-echelon form.Back-substitutionnow gives-4x3=-12 =x3= 3x2-x3=-1 =x2= 2x1+x2+ 2x3= 9 =x1= 1As the Hessian2f(x*) is positive definite (eigenvalues 1,1,4), the stationarypointx*= [ 123 ]Tis a strict local minimizer off.Moreover as the Hessian does not depend on the variablesxit is positive definite forallxR3, sofis strictly convex onR3. Hencex*is the unique global minimizeroffoverR3.Note that, for the 3 by 3 matrixG=2f(x),det (G-λI) =2-λ1112-λ1112-λ=-λ3+6λ2-9λ+4 =-(λ-1)2(λ-4) = 0gives the eigenvalues.(b)f(x) =x1(x2-1) +x3(x23-3)
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AnswerThe gradient and Hessian offaref(x) =x2-1x13x23-3,2f(x) =010100006x3The stationary points are the solutions tof(x) =0, givingx*=011,x=01-1.Atx*= [ 011 ]T, the Hessian is2f(x*) =010100006which has eigenvalues-1,1,6, so is indefinite. Hencex*is a saddle point.Atx= [ 01-1 ]T, the Hessian is2f(x) =01010000-6which has eigenvalues-6,-1,1, so is indefinite. Hencexis a saddle point.(c)f(x) = (x21+ 2x22)e-(x21+x22)Answer

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