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Solution to Problem Set 4November 9, 20131 (i)f(x) = x3− 12xf0(x) = 3x2− 12 = 0, x = 2 or − 2(ii)f(x, y) = x2− 2xy + y3− yf1= 2x − 2y = 0f2= −2x + 3y2− 1 = 0x = 1, y = 1 or x = −13, y = −132 (i)f(x) = x3− 3x, f0(x) = 3x2− 3 = 0, x = 1 or − 1f00(x) = 6xFor x = 1, f00(1) > 0, local minimumFor x = −1, f00(−1) < 0, local maximum.(ii)f(x1, x2) = x21+ x32+ 2x1x2f1= 2x1+ 2x2= 0, f2= 3x22+ 2x1= 0x1= 0, x2= 0 or x1= −23, x2=23f11= 2 > 0, f22= 6x2, f12= 2For x1= x2= 0, f11f22= 0 6> f212, no conclusion.For x1= −23, x2=23, f11f22= 2 × 4 = 8 > 4 = f212, local minimum.3 Lagrangian function:L = −x21+ x2− 3x22− λ(x1+ 2x2− 5)First order conditions:∂L∂x1= −2x1− λ = 01∂L∂x2= 1 − 6x2− 2λ = 0x1+ 2x2− 5 = 0x1= 2, x2=32, λ =
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