What is the extreme f(x,y)=x^2+4y^2-6x+8y+2 ?

The extreme f(x,y)=x^2+4y^2-6x+8y+2 is Minimum(3,-1)

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extreme f(x,y)=x^2+4y^2-6x+8y+2

extreme points

f (x, y) = x^{2} + 4 y^{2} - 6 x + 8 y + 2

Minimum (3, - 1)

Solution steps

Find the critical points:

(3, - 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 8

D (x, y) = 16

Check the sign of at each critical point

Check critical point

(3, - 1) :

Minimum

Minimum (3, - 1)

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

(x + 3)^{2} + 1

y = \frac{x ^{2} - 2 x + 1}{x ^{2} + 3}

f (x) = (7 + x) (11 - 3 x)^{\frac{1}{3}}

y = x^{- 4} \cdot e^{x^{4}}

What is the extreme f(x,y)=x^2+4y^2-6x+8y+2 ?
The extreme f(x,y)=x^2+4y^2-6x+8y+2 is Minimum(3,-1)