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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...

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...