course MTH 272 05/12,2300 assignment #030030.
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21:50:25 Query problem 7.5.10 extrema of x^2+6xy+10y^2-4y+4
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RESPONSE --> extrema of x^2+6xy+10y^2-4y+4 fx(x,y)=2x + 6y fy(x,y)=6x + 20y - 4 2x + 6y=0 2x=-6y x=-3y 6(-3y) + 20y - 4=0 -18y + 20y=4 2y=4 y=2 x=-3(2) x=-6 (-6,2) confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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21:54:06 List the relative extrema and the saddle points of the function and tell which is which, and how you obtained each.
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RESPONSE --> extrema of x^2+6xy+10y^2-4y+4 fx(x,y)=2x + 6y fy(x,y)=6x + 20y - 4 2x + 6y=0 2x=-6y x=-3y 6(-3y) + 20y - 4=0 -18y + 20y=4 2y=4 y=2 x=-3(2) x=-6 (-6,2) fxx=2 fyy=20 fxy=6 d=(2)(20) - (6)^2 40 - 36 4 d>0, fxx>0 RELATIVE MINIMUM (-6,2,0) z=(-6)^2 + (6)(-6)(2) + 10(2^2) - 4(2) + 4 z=0 confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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21:55:34 What are the critical points and what equations did you solve to get them?
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RESPONSE --> Critical Point: (-6,2,0) fx=2x + 6y and fy=6x + 20y - 4 were both set equal to zero and each variable was solved. To find the z value, the x and y values were placed into the original equation. confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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21:55:58 How did you test each critical point to determine if it is a relative max, relative min or saddle point?
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RESPONSE --> d=(2)(20) - (6)^2 40 - 36 4 d>0, fxx>0 RELATIVE MINIMUM (-6,2,0) confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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22:02:48 Query problem 7.5.28 extrema of x^3+y^3 -3x^2+6y^2+3x+12y+7
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RESPONSE --> extrema of x^3+y^3 -3x^2+6y^2+3x+12y+7 fx=3x^2 - 6x + 3 3(x^2 -2x + 1) 3(x-1)^2 (x -1)=0 x=1 fy=3y^2 + 12y + 12 3(y^2 + 4y + 4) 3(y+2)^2 (y+2)=0 y=-2 z=(1)^3 + (-2)^3 - 3(1^2) + 6(-2^2) + 3(1) + 12(-2) + 7 z=0 Critical Point:(1,-2,0) fxx=6x - 6 =6(1) - 6=0 fyy=6y + 12=6(-2) + 12 =0 fxy=0 d=0,fxx=0.....No Additional information learned confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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22:03:03 List the relative extrema and the saddle points of the function and tell which is which, and how you obtained each.
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RESPONSE --> extrema of x^3+y^3 -3x^2+6y^2+3x+12y+7 fx=3x^2 - 6x + 3 3(x^2 -2x + 1) 3(x-1)^2 (x -1)=0 x=1 fy=3y^2 + 12y + 12 3(y^2 + 4y + 4) 3(y+2)^2 (y+2)=0 y=-2 z=(1)^3 + (-2)^3 - 3(1^2) + 6(-2^2) + 3(1) + 12(-2) + 7 z=0 Critical Point:(1,-2,0) fxx=6x - 6 =6(1) - 6=0 fyy=6y + 12=6(-2) + 12 =0 fxy=0 d=0,fxx=0.....No Additional information learned confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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22:03:19 What are the critical points and what equations did you solve to get them?
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RESPONSE --> extrema of x^3+y^3 -3x^2+6y^2+3x+12y+7 fx=3x^2 - 6x + 3 3(x^2 -2x + 1) 3(x-1)^2 (x -1)=0 x=1 fy=3y^2 + 12y + 12 3(y^2 + 4y + 4) 3(y+2)^2 (y+2)=0 y=-2 z=(1)^3 + (-2)^3 - 3(1^2) + 6(-2^2) + 3(1) + 12(-2) + 7 z=0 Critical Point:(1,-2,0) confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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22:03:33 How did you test each critical point to determine if it is a relative max, relative min or saddle point?
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RESPONSE --> Critical Point:(1,-2,0) fxx=6x - 6 =6(1) - 6=0 fyy=6y + 12=6(-2) + 12 =0 fxy=0 d=0,fxx=0.....No Additional information learned confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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22:04:05 At what point(s) did the second-partials test fail?
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RESPONSE --> Second partials failed for (1,-2,0). confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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