ha lvh262 Homework 14 7 karakurt 56295 This print out should have 26 questions Multiple choice questions may continue on the next column or page find all choices before answering 001 10 0 points Determine the absolute maximum value of 1 the only critical points inside the square occur when cos x 0 cos y 0 i e at 2 2 and at this critical point 0 f 2 2 We look at the edges separately f x y 2 cos x cos y i on Edge 1 f x 0 2 cos x so on the square 0 x y max f x 0 2 0 x 1 absolute max 2 i on Edge 2 f 0 y 2 cos y so 2 absolute max 0 max f 0 y 2 0 y 3 absolute max 2 correct i on Edge 3 f x 2 cos x so 4 absolute max 2 max f x 2 5 absolute max 2 0 x Explanation The absolute extrema of f x y occur either at a critical point inside the square or on the boundary of the square as shown in i on Edge 4 f y 2 cos x so max f y 2 0 y Consequently on the square f has y edge 3 edge 4 edge 2 absolute maximum 2 keywords x 10 0 points f x y x2 y 2 x y 1 Now on the unit disk f 2 cos x sin y y Thus the critical points are at the solutions of 2 sin x cos y 0 002 Determine the absolute maximum of edge 1 f 2 sin x cos y x 2 cos x sin y 0 Since 0 x y sin x sin y 0 D x y x2 y 2 1 1 absolute max 2 2 2 correct 2 absolute max 3 3 absolute max 2 ha lvh262 Homework 14 7 karakurt 56295 2 Consequently on D f x y has 1 4 absolute max 2 Absolute max 2 2 5 absolute max 2 Explanation The Absolute Extrema occur either at a critical point inside the disk or on the boundary of the disk 1 Inside D when 2 2 f x y x y x y 1 the critical points occur at the solutions to f 2x 1 x f 2y 1 0 y So 1 2 1 2 is the only critical point and this lies inside the disk At this critical point 1 1 1 f 2 2 2 2 On the boundary of D the boundary of D is the unit circle which can be parametrized by r t cos t i sin t j 0 t 2 Thus the restriction of f to the boundary is the function f r t 2 cos t sin t Finding the absolute extrema of f r t on 0 2 is a single variable problem they will occur either at a critical point in 0 2 or at the endpoints t 0 2 But df r t sin t cos t dt so the critical points occur when tan t 1 But 2 2 f r 4 of f r t on 0 2 i e at t 4 5 4 5 f r 2 2 4 keywords 003 10 0 points Suppose 1 1 is a critical point of a function f having continuous second derivatives such that fxx 1 1 4 fxy 1 1 2 fyy 1 1 5 Which of the following properties does f have at 1 1 1 a local maximum 2 a local minimum correct 3 a saddle point Explanation Since 1 1 is a critical point the Second Derivative test ensures that f will have 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 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 iii a saddle point at 1 1 if fxx 1 1 fyy 1 1 fxy 1 1 2 From the given values of the second derivatives of f at 1 1 it thus follows that f has while f r 0 1 f r 2 a local minimum ha lvh262 Homework 14 7 karakurt 56295 at 1 1 004 B TRUE 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 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 P Q 0 C FALSE the point Q lies on the 0contour 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 keywords contour map local extrema True False 1 2 3 005 R 10 0 points Locate and classify the local extremum of the function 2 1 0 1 2 3 A local minimum at P f x y x2 y 2 8x 6y 2 1 saddle point at 4 3 B local maximum at R 2 local maximum at 4 3 C local maximum at Q 3 saddle point at 4 4 1 A and C only 4 local minimum at 4 3 correct 2 A only 5 local maximum at 3 3 3 none of them 4 all of them 5 A and B only correct 6 C only 7 B and C only 8 B only Explanation A 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 Explanation Differentiating the function with respect to x and y we see that fx 2x 8 fy 2y 6 As the local extremum is a critical point therefore the local extremum will occur at 4 3 To classify the critical point we use the second derivative test Now fxx fyy 2 while fxy 0 Thus fxx fyy 0 fxx fyy fxy 2 0 Hence f has a local minimum at 4 3 ha lvh262 Homework 14 7 karakurt 56295 4 Locate and classify all the local extrema of 006 10 0 points f x y 2x3 2y 3 6xy 4 Locate and classify the local extremum of the function f x y 4 8x 2y x2 2y 2 1 local max at 1 1 saddle at 0 0 2 saddle at 1 1 local max at 0 0 1 saddle point at 1 2 3 local max at 1 1 saddle at 0 0 correct 2 local max at 1 2 1 3 saddle point at 4 2 1 4 local min at 4 2 1 correct 5 local max at 4 2 4 local min at 0 0 local max at 1 1 5 local min at 1 1 saddle at 0 0 Explanation Local …