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

The extreme f(x,y)=2x^2+5y^2-xy is Minimum(0,0)

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

extreme points

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

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 10

D (x, y) = 39

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

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

y = \frac{1}{3} x^{3} + \frac{1}{2} x^{2} - 6 x + 8

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

y = (x^{2} - 9)^{4}

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

What is the extreme f(x,y)=2x^2+5y^2-xy ?
The extreme f(x,y)=2x^2+5y^2-xy is Minimum(0,0)