What is the extreme f(x,y)=x^2+3xy+4.5y^2+4x ?

The extreme f(x,y)=x^2+3xy+4.5y^2+4x is Minimum(-4, 4/3)

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extreme f(x,y)=x^2+3xy+4.5y^2+4x

extreme points

f (x, y) = x^{2} + 3 x y + 4.5 y^{2} + 4 x

Minimum (- 4, \frac{4}{3})

Solution steps

Find the critical points:

(- 4, \frac{4}{3})

\frac{\partial ^{2} f}{\partial x ^{2}} = 9

D (x, y) = 9

Check the sign of at each critical point

Check critical point

(- 4, \frac{4}{3}) :

Minimum

Minimum (- 4, \frac{4}{3})

5 x^{3} - 3 x^{5}

S (u, t) = \frac{u + u ^{0}}{2} \cdot t

f (x) = (x^{2} + 1) e^{x}

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

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

What is the extreme f(x,y)=x^2+3xy+4.5y^2+4x ?
The extreme f(x,y)=x^2+3xy+4.5y^2+4x is Minimum(-4, 4/3)