UT M 408C - HW 14.7-solutions-1 (11 pages)

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HW 14.7-solutions-1



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lewis scl876 HW 14 7 radin 55315 This print out should have 17 questions Multiple choice questions may continue on the next column or page find all choices before answering 001 10 0 points Suppose 1 1 is a critical point of a function f having continuous second derivatives such that 1 at 1 1 002 10 0 points In the contour map below identify the points P Q and R as local minima local maxima or neither 3 2 fxx 1 1 5 fxy 1 1 2 1 0 1 fyy 1 1 5 2 P Q Which of the following properties does f have at 1 1 0 R 2 1 a saddle point correct 1 0 1 2 a local maximum 2 3 3 a local minimum Explanation Since 1 1 is a critical point the Second Derivative test ensures that f will have A local minimum at R B local maximum at P C neither local max nor local min at Q i a local minimum at 1 1 if fxx 1 1 0 fyy 1 1 0 fxx 1 1 fyy 1 1 fxy 1 1 2 1 A only 2 B only 3 C only correct ii a local maximum at 1 1 if fxx 1 1 0 fyy 1 1 0 fxx 1 1 fyy 1 1 fxy 1 1 2 4 A and C only 5 A and B only 6 all of them iii a saddle point at 1 1 if 7 B and C only 2 fxx 1 1 fyy 1 1 fxy 1 1 From the given values of the second derivatives of f at 1 1 it thus follows that f has a saddle point 8 none of them Explanation A 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 lewis scl876 HW 14 7 radin 55315 B FALSE 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 not a local maximum C TRUE the point Q lies on the 0 contour and in 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 has neither a local maximum nor a local minimum at Q keywords contour map local extrema True False 003 part 1 of 3 10 0 points If f is defined by 8 f x y x3 x2 4xy 2y 2 y 3 locate the critical points of f 1 3 1 0 1 4 2 4 1 1 3 2 0 4 2 4 1 3 1 3 0 4 2 4 1 1 3 4 0 4 2 4 1 3 1 5 0 correct 4 2 4 1 1 3 6 0 4 2 4 Explanation After differentiation 2 Eliminating y we see that 8x2 2x 1 1 1 8 x x 0 4 2 1 1 the solutions of which are x and 4 2 Consequently the critical points of f occur at 1 0 4 3 1 2 4 004 part 2 of 3 10 0 points Which of the following most accurately describes the behaviour of f 1 1 local minimum at 0 4 1 2 local maximum at 0 4 1 3 saddle point at 0 correct 4 1 4 saddle point at 0 4 1 5 local minimum at 0 4 1 6 local maximum at 0 4 Explanation After differentiation once again we see that fx 8x2 2x 4y fxx 2 8x 1 fyy 4 fxy 4 while fy 4x 4y 1 Thus the critical points of f are the common solutions of the equations 8x2 2x 4y 0 4x 4y 1 0 and so at the critical point 1 0 4 A fxx 1 2 4 0 lewis scl876 HW 14 7 radin 55315 and 3 and so at the critical point C fyy 4 0 14 0 A fxx 1 3 2 4 while But then B fxy 1 4 4 0 1 saddle point at 0 4 005 part 3 of 3 10 0 points Which of the following most accurately describes the behaviour of f 3 1 1 saddle point at 2 4 1 3 2 local maximum at 2 4 1 3 3 local minimum at correct 2 4 1 3 4 local minimum at 2 4 1 3 5 local maximum at 2 4 1 3 6 saddle point at 2 4 1 3 7 local minimum at 2 4 1 3 8 local maximum at 2 4 1 3 9 saddle point at 2 4 Explanation After differentiation once again we see that fxx 2 8x 1 fyy 4 fxy 4 3 1 2 4 10 and C fyy 1 3 2 4 AC B 2 24 0 Hence by the second derivative test f has a 4 0 while B fxy 1 3 2 4 4 But then AC B 2 24 0 Hence by the second derivative test f has a local minimum at 006 3 1 2 4 10 0 points Locate and classify the local extremum of f when f x y 3x 1 y 2 3 xy x y 0 1 local min at 3 3 2 local min at 1 3 3 3 saddle at 3 3 4 local max at 3 3 1 5 saddle at 3 3 1 3 6 local max at 3 Explanation correct lewis scl876 HW 14 7 radin 55315 4 Differentiating once we see that 1 3 2 x y fx 1 1 fy 2 3 xy At a local extremum these first partial derivatives are zero Thus f has a local extremum 1 at 3 3 To classify the local extremum we use the second derivative test Now fxx 2 1 2 fxy 2 2 fyy 3 3 x y x y xy But then A fxx 1 18 0 3 3 2 0 9 3 3 C fyy 1 and Hence by the second derivative test f has a local min at 007 1 3 3 fx 6 x2 y fy 6 y 2 x The critical points of f are thus the common solutions of the equations x2 y 0 y 2 x 0 This yields only the two extremum points 1 1 and 0 0 But after differentiating again we see that fxx 12x fxy 6 fyy 12y consequently fxxfyy fxy 2 12 2 xy 36 local max at 1 1 saddle at 0 0 AC B 2 3 0 Explanation Local extrema occur at the critical points of f Now after differentiation of f we obtain Hence by the second derivative test there are 1 3 3 B fxy 1 Thus 5 local max at 1 1 saddle at 0 0 10 0 points Locate and classify all the local extrema of f x y 2x3 2y 3 6xy 1 1 local max at 1 …


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