What is the extreme f(x,y)=2x^2+3y^2-4x-3y-10 ?

The extreme f(x,y)=2x^2+3y^2-4x-3y-10 is Minimum(1, 1/2)

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extreme f(x,y)=2x^2+3y^2-4x-3y-10

extreme points

f (x, y) = 2 x^{2} + 3 y^{2} - 4 x - 3 y - 10

Minimum (1, \frac{1}{2})

Solution steps

Find the critical points:

(1, \frac{1}{2})

\frac{\partial ^{2} f}{\partial x ^{2}} = 6

D (x, y) = 24

Check the sign of at each critical point

Check critical point

(1, \frac{1}{2}) :

Minimum

Minimum (1, \frac{1}{2})

f (x, y) = e^{y} - y e^{x}

f (x) = 2 x + cos (2 x), 0 < x < π

f (x, y) = x^{2} - 3 x y + 5 x - 2 y + 6 y^{2} + 8

f (x) = \frac{3}{4} x^{4} - 3 x^{3} - 81 x^{2} - 16

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

What is the extreme f(x,y)=2x^2+3y^2-4x-3y-10 ?
The extreme f(x,y)=2x^2+3y^2-4x-3y-10 is Minimum(1, 1/2)