determine the position and nature of the stationary points of the following

[waco3002]

determine the position and nature of the stationary points of the following of the function z(x,y) = x^2 + y^3 - 4xy + 4?


so far i have

pdz/pdx = 2x -4y =0 .....x=2y
pdz/pdy = 3y^2-4x=0
3y^2 -4(2y)=0
3y^2-8y=0
y(3y-8)=0
y=0, 8/3
x= 0,16/3

(0,0,) (16/3,8/3)


p2dz/pdx^2 = 2
p2dz/dy^2 = 6y
p2dz/pdxpdy= p2dz/dydx= -16

how do i find the saddle point, maximum and minimum points?

 

[Ishuda]

determine the position and nature of the stationary points of the following of the function z(x,y) = x^2 + y^3 - 4xy + 4?


so far i have

pdz/pdx = 2x -4y =0 .....x=2y
pdz/pdy = 3y^2-4x=0
3y^2 -4(2y)=0
3y^2-8y=0
y(3y-8)=0
y=0, 8/3
x= 0,16/3

(0,0,) (16/3,8/3)


p2dz/pdx^2 = 2
p2dz/dy^2 = 6y
p2dz/pdxpdy= p2dz/dydx= -16 <===incorrect

how do i find the saddle point, maximum and minimum points?

See above
\(\displaystyle \frac{\partial^2 z}{\partial x\, \partial y}=-4\)

What is the definition of a minimum/maximum/saddle point. You might start at
http://mathworld.wolfram.com/SecondDerivativeTest.html
or
http://tutorial.math.lamar.edu/Classes/CalcIII/RelativeExtrema.aspx

 

[waco3002]

the critical points are at (0, 0) and (16/3, 8/3).

the Second Derivative Test.

z_xx = 2,
z_yy = 6y,
z_xy = -4

==> D = (z_xx)(z_yy) - (z_xy) = 12y - 16.

Since D(0, 0) = -16 < 0, we have a saddle point at (0, 0).

Since D(16/3, 8/3) = 16 > 0 and z_xx (16/3, 8/3) = 2 > 0, we have a local
minimum at (16/3, 8/3)


​does this look correct?

 

[Ishuda]

the critical points are at (0, 0) and (16/3, 8/3).

the Second Derivative Test.

z_xx = 2,
z_yy = 6y,
z_xy = -4

==> D = (z_xx)(z_yy) - (z_xy) = 12y - 16.

Since D(0, 0) = -16 < 0, we have a saddle point at (0, 0).

Since D(16/3, 8/3) = 16 > 0 and z_xx (16/3, 8/3) = 2 > 0, we have a local
minimum at (16/3, 8/3)


​does this look correct?

Looks good to me