What is the extreme f(x,y)= 1/3 x^3+2y^2-3x^2+5x-8y+4 ?

The extreme f(x,y)= 1/3 x^3+2y^2-3x^2+5x-8y+4 is Saddle(1,2),Minimum(5,2)

Popular >

extreme f(x,y)= 1/3 x^3+2y^2-3x^2+5x-8y+4

extreme points

f (x, y) = \frac{1}{3} x^{3} + 2 y^{2} - 3 x^{2} + 5 x - 8 y + 4

Saddle (1, 2), Minimum (5, 2)

Solution steps

Find the critical points:

(1, 2), (5, 2)

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

D (x, y) = 4 (2 x - 6)

Check the sign of at each critical point

Check critical point

(1, 2) :

Saddle

Check critical point

(5, 2) :

Minimum

Saddle (1, 2), Minimum (5, 2)

f (x) = x^{2} y - x y^{2}

g (x, y) = x^{2} - 3 x y - y^{2}

f (x, y) = x^{3} + x y + y^{2}

f (x) = x^{2} - 2 x - 3

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

What is the extreme f(x,y)= 1/3 x^3+2y^2-3x^2+5x-8y+4 ?
The extreme f(x,y)= 1/3 x^3+2y^2-3x^2+5x-8y+4 is Saddle(1,2),Minimum(5,2)