What is the extreme f(x,y)=x^2+4y^2-6x+16y ?

The extreme f(x,y)=x^2+4y^2-6x+16y is Minimum(3,-2)

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extreme f(x,y)=x^2+4y^2-6x+16y

extreme points

f (x, y) = x^{2} + 4 y^{2} - 6 x + 16 y

Minimum (3, - 2)

Solution steps

Find the critical points:

(3, - 2)

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

D (x, y) = 16

Check the sign of at each critical point

Check critical point

(3, - 2) :

Minimum

Minimum (3, - 2)

f (x, y) = 8 x y - x^{3} - 4 y^{2}

f (x) = 5 x^{6} - 3 x^{4} + 2 x - 9

f (x, y) = x^{3} + y^{3} - 9 x y + 27

f (x) = 6 x^{2} ln (x)

x^{4} - 4 x^{3} + 1

What is the extreme f(x,y)=x^2+4y^2-6x+16y ?
The extreme f(x,y)=x^2+4y^2-6x+16y is Minimum(3,-2)