What is the extreme f(x,y)=x^2+2y^2+3x+10 ?

The extreme f(x,y)=x^2+2y^2+3x+10 is Minimum(-3/2 ,0)

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extreme f(x,y)=x^2+2y^2+3x+10

extreme

Minimum (- \frac{3}{2}, 0)

Solution steps

Find the critical points:

(- \frac{3}{2}, 0)

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

D (x, y) = 8

Check the sign of at each critical point

Check critical point

(- \frac{3}{2}, 0) :

Minimum

Minimum (- \frac{3}{2}, 0)

What is the extreme f(x,y)=x^2+2y^2+3x+10 ?
The extreme f(x,y)=x^2+2y^2+3x+10 is Minimum(-3/2 ,0)