Answer:
A(-1,0) is a local maximum point.
B(-1,0) is a saddle point
C(3,0) is a saddle point
D(3,2) is a local minimum point.
Step-by-step explanation:
The given function is
[tex]f(x,y)=x^3+y^3-3x^2-3y^2-9x[/tex]
The first partial derivative with respect to x is
[tex]f_x=3x^2-6x-9[/tex]
The first partial derivative with respect to y is
[tex]f_y=3y^2-6y[/tex]
We now set each equation to zero to obtain the system of equations;
[tex]3x^2-6x-9=0[/tex]
[tex]3y^2-6y=0[/tex]
Solving the two equations simultaneously, gives;
[tex]x=-1,x=3[/tex] and [tex]y=0,y=2[/tex]
The critical points are
A(-1,0), B(-1,2),C(3,0),and D(3,2).
Now, we need to calculate the discriminant,
[tex]D=f_{xx}(x,y)f_{yy}(x,y)-(f_{xy}(x,y))^2[/tex]
But, we would have to calculate the second partial derivatives first.
[tex]f_{xx}=6x-6[/tex]
[tex]f_{yy}=6y-6[/tex]
[tex]f_{xy}=0[/tex]
[tex]\Rightarrow D=(6x-6)(6y-6)-0^2[/tex]
[tex]\Rightarrow D=(6x-6)(6y-6)[/tex]
At A(-1,0),
[tex]D=(6(-1)-6)(6(0)-6)=72\:>\:0[/tex] and [tex]f_{xx}=6(-1)-6=-18\:<\:0[/tex]
Hence A(-1,0) is a local maximum point.
See graph
At B(-1,2);
[tex]D=(6(-1)-6)(6(2)-6)=-72\:<\:0[/tex]
Hence, B(-1,0) is neither a local maximum or a local minimum point.
This is a saddle point.
At C(3,0)
[tex]D=(6(3)-6)(6(0)-6)=-72\:<\:0[/tex]
Hence, C(3,0) is neither a local minimum or maximum point. It is a saddle point.
At D(3,2),
[tex]D=(6(3)-6)(6(2)-6)=72\:>\:0[/tex] and [tex]f_{xx}=6(3)-6=12\:>\:0[/tex]
Hence D(3,2) is a local minimum point.
See graph in attachment.
Answer:
A(-1,0) is a local maximum point.
B(-1,2) is saddle point.
C(3,0) is also a saddle point
D(3,2) is the minimum point.
Step-by-step explanation:
Given information:
The function : [tex]f(x,y)=x^3=y^3-3x^2-3y^2-9x[/tex]
The first partial derivative with respect to x is:
[tex]f'_x=3x^2-6x-9[/tex]
And the partial derivative with respect to y is:
[tex]f'_y=3y^2-6y[/tex]
Now equate the above derivatives to zero, to obtain the system of equation:
[tex]3x^2-6x-9=0[/tex]
[tex]3y^2-6y=0[/tex]
On solving the two equations:
The critical points are : A(-1,0), B(-1,-2), C(3,0) and D(3,2)
Now calculate the discriminant
[tex]D=f{xx}(x,y)f_{yy}(x,y)-f_{xy}(x,y)^2[/tex]
Here,
[tex]f_{xx}=6x-6\\f_{yy}=6y-6\\f_{xy}=0\\[/tex]
On putting the above values we get:
[tex]D=(6x-6)(6y-6)[/tex]
At, A(-1,0)
[tex]D=-18<0[/tex]
Hence , A(-1,0) is a local maximum point.
At, B(-1,2)
[tex]D=-72<0\\[/tex]
Hence B(-1,2) is saddle point.
At, C(3,0)
[tex]D=-72[/tex]
Hence C(3,0) is also a saddle point
At, D(3,2)
[tex]D=12>0[/tex]
Hence , D(3,2) is the minimum point.
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