What is the extreme f(x,y)=5x^2-4xy+2y^2-36x ?

The extreme f(x,y)=5x^2-4xy+2y^2-36x is Minimum(6,6)

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extreme f(x,y)=5x^2-4xy+2y^2-36x

extreme points

f (x, y) = 5 x^{2} - 4 x y + 2 y^{2} - 36 x

Minimum (6, 6)

Solution steps

Find the critical points:

(6, 6)

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

D (x, y) = 24

Check the sign of at each critical point

Check critical point

(6, 6) :

Minimum

Minimum (6, 6)

f (x, y) = 37 - x^{2} - y^{2}

f (x) = 0.09 x^{2} + 16 x + 350, 0 < x < 150

f (x) = - x^{2} + 5 x - 6, 2 \leq x \leq 3

f (x, y) = x^{3} - y^{2} - 6 x y - 4

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

What is the extreme f(x,y)=5x^2-4xy+2y^2-36x ?
The extreme f(x,y)=5x^2-4xy+2y^2-36x is Minimum(6,6)