22_new - HW10 gilbert (55035) This print-out should have 16...

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HW10 – gilbert – (55035) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points In the contour map below identify the points P, Q, and R as local minima, local maxima, or neither. 3 2 1 0 -1 -2 P Q 0 R 2 1 0 -1 -2 -3 A. local maximum at Q, B. local minimum at R, C. local minimum at P . 1. A and C only 2. A and B only 3. B and C only 4. C only correct 5. none of them 6. A only 7. B only 8. all of them Explanation: A. FALSE: the point Q lies on the 0- contour and this contour divides the region near Q into two regions. In one region the contours have values increasing to 0, while in the other the contours have values decreasing to 0. So the surface does not have a local minimum at Q. B. FALSE: the contours near R are closed curves enclosing R and the contours increase in value as we approch R. So the surface has a local maximum at R, not a local minimum. C. TRUE: the contours near P are closed curves enclosing P and the contours decrease in value as we approch P . So the surface has a local minimum at P . keywords: contour map, local extrema, True/False, 002 10.0 points Locate and classify all the local extrema of f (x, y) = x 3 − y 3 + 3xy − 2 . 1. local max at (1, −1), local min at (0, 0) 2. local max at (1, −1), saddle point at (0, 0) 3. local min at (0, 0), saddle point at (1, −1) 4. local min at (1, −1), saddle point at (0, 0) correct 5. local max at (0, 0), saddle point at (1, −1) Explanation: Since f has derivatives everywhere, the crit-ical points occur at the solutions of f (x, y) = f X i + f Y j = 0 . But f X = 0 when ∂f ∂x = 3x 2 + 3y = 0 , i.e., y = −x 2 ,
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HW10 – gilbert – (55035) 2 while f Y = 0 when ∂f ∂y = −3y 2 + 3x = 0 , i.e., x = y 2 . Substituting the first into the second yields x = (−x 2 ) 2 = x 4 , which can be written as x(1 − x 3 ) = 0 , i.e., x = 0,1 . Thus, the critical points are (0, 0), (1, −1) . and to classify these critical points we use the Second Derivative test. Now f XX = 6x , f YY = −6y , f XY = 3 . But then, at (0, 0), A = f XX (0, 0) = 0, B = f XY (0, 0) = 3 , while C = f YY (0, 0) = 0 . Consequently, D = AC − B 2 = −9 < 0 . and so there is a saddle point at (0, 0) . On the other hand, at (1, −1), A = f XX (1, −1) = 6, B = f XY (1, −1) = 3 , while C = f YY (1, −1) = 6 . Thus D = AC − B 2 = 27 > 0 . and so, since A, C > 0, there is a local minimum at (1, −1) . keywords: 003 (part 1 of 3) 10.0 points If f is defined by f (x, y) = 4 3 x 3 + x 2 + 4xy + 2y 2 + 2y , (i) locate the critical points of f . 1. 2 1 , 0 , −1, 2 3 2.
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22_new - HW10 gilbert (55035) This print-out should have 16...

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