What is the extreme f(x,y)=4x^2y+2xy^2-12xy-5 ?

The extreme f(x,y)=4x^2y+2xy^2-12xy-5 is Minimum(1,2),Saddle(0,6)

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extreme f(x,y)=4x^2y+2xy^2-12xy-5

extreme points

f (x, y) = 4 x^{2} y + 2 x y^{2} - 12 x y - 5

Minimum (1, 2), Saddle (0, 6)

Solution steps

Find the critical points:

(1, 2), (0, 6)

\frac{\partial ^{2} f}{\partial x ^{2}} = 4 x

D (x, y) = - 64 x^{2} - 32 x y + 192 x - 16 y^{2} + 96 y - 144

Check the sign of at each critical point

Check critical point

(1, 2) :

Minimum

Check critical point

(0, 6) :

Saddle

Minimum (1, 2), Saddle (0, 6)

f (x) = x^{3} + \frac{48}{x}

f (x, y) = \frac{y ^{2}}{y + x ^{2}}

f (x) = \frac{1}{3} x^{3} + \frac{1}{2} x^{2} - 6 x + 8

f (x) = x^{\frac{2}{3}}

f (x, y) = 9 - 2 x + 4 y - x^{2} - 4 y^{2}

What is the extreme f(x,y)=4x^2y+2xy^2-12xy-5 ?
The extreme f(x,y)=4x^2y+2xy^2-12xy-5 is Minimum(1,2),Saddle(0,6)