What is the extreme f(x,y)=4x^2+3xy+5y^2+5y+3y ?

The extreme f(x,y)=4x^2+3xy+5y^2+5y+3y is Minimum(24/71 ,-64/71)

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extreme f(x,y)=4x^2+3xy+5y^2+5y+3y

extreme points

f (x, y) = 4 x^{2} + 3 x y + 5 y^{2} + 5 y + 3 y

Minimum (\frac{24}{71}, - \frac{64}{71})

Solution steps

Find the critical points:

(\frac{24}{71}, - \frac{64}{71})

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

D (x, y) = 71

Check the sign of at each critical point

Check critical point

(\frac{24}{71}, - \frac{64}{71}) :

Minimum

Minimum (\frac{24}{71}, - \frac{64}{71})

f (x) = \frac{( 5 x ^{2} )}{ln ( x )}

f (x) = x^{3} - 2 x^{2} + 14

f (x) = x^{3} - 2 x^{2} + 10

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

x^{3} - 3 x^{2} - 5 x - 5

What is the extreme f(x,y)=4x^2+3xy+5y^2+5y+3y ?
The extreme f(x,y)=4x^2+3xy+5y^2+5y+3y is Minimum(24/71 ,-64/71)