What is the extreme f(x)=3x^2+2y^2-xy-4x-7y+12 ?

The extreme f(x)=3x^2+2y^2-xy-4x-7y+12 is Minimum(1,2)

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extreme f(x)=3x^2+2y^2-xy-4x-7y+12

extreme points

f (x) = 3 x^{2} + 2 y^{2} - x y - 4 x - 7 y + 12

Minimum (1, 2)

Solution steps

Find the critical points:

(1, 2)

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

D (x, y) = 23

Check the sign of at each critical point

Check critical point

(1, 2) :

Minimum

Minimum (1, 2)

y = \frac{\frac{25}{32} x ^{3} + 6400}{x}

f (x) = x ∣ x ∣

E (x) = x^{4} - 8 x + 3 x^{2} + x m - 5 - 11

f (x, y) = y^{3} + (6 x^{2}) y - 6 x^{2} - 6 y^{2} + 3

x^{5}

What is the extreme f(x)=3x^2+2y^2-xy-4x-7y+12 ?
The extreme f(x)=3x^2+2y^2-xy-4x-7y+12 is Minimum(1,2)