Page 1 Math 251 copyright Joe Kahlig 15C 1 a False notice fx and fy are nice polynomials in x and y Thus fxy and fyx should be equal but they are not b True j h0 1 0i c False Notice D 0 thus saddle point d True 2 a g 1 2 3 h18 12 12i 1 b u h2 1 2i 3 2 1 2 Du g 1 2 3 18 12 12 16 3 3 3 3 path gives f x 5x 1 lim f x 5x 1 x 1 6x2 5x 1 2x 2 7 By L Hopital method 2 4 the normal vector is hfx fy 1i at the point n h15 10 1i tangent plane 15 x 2 10 y 1 1 z 15 0 or 15x 10y z 25 5 a f 2 1 h4 2i h4 2i 20 b h 4 2i 6 since we know f 2 1 3 then dx x 1 7 2 0 3 dy y 2 2 p 1 1 2 let z f x y 4 x2 y 2 x fx p and fx 2 1 4 x2 y 2 y fy p and fy 2 1 4 x2 y 2 2 3 1 3 dz fx dx fy dy thus dz 0 2 f 1 7 2 2 f 2 1 dz 3 0 2 3 2 7 Let f x y z 2x3 y 3 z 5 11 Then n f 1 2 1 h6 12 5i parametric equation of the normal line x 1 6t y 2 12t z 1 5t 8 ga fx tat 1 fy 2e2a gt fx at ln a 4t3 fy 3t2 sec2 t3 Page 2 Math 251 copyright Joe Kahlig 15C 9 left f x2 y 2 z 2 1 and g 2x y 3z 2 we want f k g solving gets two possible points 10 zx 11 x 2 1 3 14 14 14 or 2 1 3 14 14 14 2x sin x3 y 2 3x4 cos x3 y 2 Fx Fz 2yz 4 sin 4z 1 1 3 y z 25 50 50 12 fx x2 4y 9 and fy 4x 2y solving fy 0 gives y 2x now sub this into fx 0 and get x2 8x 9 0 after factoring x 9 and x 0 Critical points 1 2 saddle point 9 18 local max 13 step 1 find fx and fy and solve for any critical points in the interior you get the point 2 2 note this in not in the region so ignore it Boundary 1 x 4 line x 4 and y y for 2 y 2 g y f 4 y 16 8y y 2 g 0 8 2y set equal to zero and solve has a critical value at y 4 note this is outside the region so ignore it Boundary 2 parabola x y 2 x y 2 y y for 2 y 2 g y f y 2 y 15y 2 2y 3 g 0 10y 6y 2 set equal to zero and solve has a critical value at y 0 and y 5 3 critical points 0 0 and 25 9 5 3 critical points 4 2 4 2 0 0 25 9 5 3 abs max 36 abs min 0 function value 36 4 0 125 27
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