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prac2Csol

prac2Csol - 18.02 Practice Exam 2 B Solutions Problem 1 a f...

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18.02 Practice Exam 2 B – Solutions Problem 1. a) f = h 2 xy 2 - 1 , 2 x 2 y i = h 3 , 8 i = 3 ˆ i - 8 ˆ j . b) z - 2 = 3( x - 2) + 8( y - 1) or z = 3 x + 8 y - 12. c) Δ x = 1 . 9 - 2 = - 1 / 10 and Δ y = 1 . 1 - 1 = 1 / 10. So z 2+3Δ x +8Δ y = 2 - 3 / 10+8 / 10 = 2 . 5 d) df ds fl fl fl fl ˆ u = f · ˆ u = h 3 , 8 i · h- 1 , 1 i 2 = - 3 + 8 2 = 5 2 Problem 2. Problem 3. a) w x = - 6 x - 4 y + 16 = 0 - 3 x - 2 y + 8 = 0 w y = - 4 x - 2 y - 12 = 0 4 x + 2 y + 12 = 0 x = - 20 y = 34 Therefore there is just one critical point at ( - 20 , 34). Since w xx w yy - w 2 xy = ( - 6) ( - 2) - ( - 4) 2 = 12 - 16 = - 4 < 0 , the critical point is a saddle point. b) There is no critical point in the first quadrant, hence the maximum must be at infinity or on the boundary of the first quadrant. The boundary is composed of two half-lines: x = 0 and y 0 on which w = - y 2 - 12 y . It has a maximum ( w = 0) at y = 0. y = 0 and x 0, where w = - 3 x 2 + 16 x . (The graph is a parabola pointing downwards). Maximum: w x = - 6 x + 16 = 0 x = 8 / 3. Hence w has a maximum at (8 / 3 , 0) and w = - 3(8 / 3) 2 + 16 · 8 / 3 = 64 / 3 > 0.

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prac2Csol - 18.02 Practice Exam 2 B Solutions Problem 1 a f...

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