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

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

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

extreme points

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

Minimum (4, - 2)

Solution steps

Find the critical points:

(4, - 2)

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

D (x, y) = 8

Check the sign of at each critical point

Check critical point

(4, - 2) :

Minimum

Minimum (4, - 2)

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

f (x, y) = 5 x + 6 y

x^{2} + 6 x y + 2 y^{2}

+ \frac{x + \frac{1}{x} + 8}{x + \frac{1}{x} + 2}

f (x, y) = 6 (1 - x^{(2)}) \cdot (1 - y^{(2)})

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