solutions for final exam - 18.02 Practice Exam 2 A...

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18.02 Practice Exam 2 A Solutions Problem 1. a) f = ( y 4 x 3 ı + x ˆ ȷ ; at P , f = ⟨− 3 , 1 . b) w ≃ − 3 x + y . Problem 2. dh h a) By measuring, h = 100 for s - 500, so - - . 2. ds u ˆ s b) Q is the northernmost point on the curve h = 2200; the vertical distance between consecutive h h 100 level curves is about 1/3 of the given length unit, so - - ≃ − . 3. y y 1000 / 3 Problem 3. f ( x, y, z ) = x 3 y + z 2 = 3 : the normal vector is f = 3 x 2 y, x 3 , 2 z - = 3 , 1 , 4 . The tangent plane is 3 x y + 4 z = 4. Problem 4. a) The volume is xyz = xy (1 x 2 y 2 ) = xy x 3 y xy 3 . Critical points: f x = y 3 x 2 y y 3 = 0, f y = x x 3 3 xy 2 = 0. b) Assuming x > 0 and y > 0, the equations can be rewritten as 1 3 x 2 y 2 = 0, 1 x 2 3 y 2 = 0. Solution: x 2 = y 2 = 1 / 4, i.e. ( x, y ) = (1 / 2 , 1 / 2). c) f xx = 6 xy = 3 / 2, f yy = 6 xy = 3 / 2, f xy = 1 3 x 2 = 1 / 2. So f xx f yy xy > 0, 2 3 y f 2 and f xx < 0, it is a local maximum.
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