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

The extreme f(x,y)=2x^2+4x+2y^2+5y is Minimum(-1,-5/4)

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

extreme points

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

Minimum (- 1, - \frac{5}{4})

Solution steps

Find the critical points:

(- 1, - \frac{5}{4})

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

D (x, y) = 16

Check the sign of at each critical point

Check critical point

(- 1, - \frac{5}{4}) :

Minimum

Minimum (- 1, - \frac{5}{4})

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

- x^{3} + 5 x^{2} + 8 x + 9

- 2 x^{2} + 4 x - 8

- 2 x^{2} + 4 x - 3

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

What is the extreme f(x,y)=2x^2+4x+2y^2+5y ?
The extreme f(x,y)=2x^2+4x+2y^2+5y is Minimum(-1,-5/4)