What is the extreme f(x,y)=2x^2-5x+6y+2y^2 ?

The extreme f(x,y)=2x^2-5x+6y+2y^2 is Minimum(5/4 ,-3/2)

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extreme f(x,y)=2x^2-5x+6y+2y^2

extreme points

f (x, y) = 2 x^{2} - 5 x + 6 y + 2 y^{2}

Minimum (\frac{5}{4}, - \frac{3}{2})

Solution steps

Find the critical points:

(\frac{5}{4}, - \frac{3}{2})

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

D (x, y) = 16

Check the sign of at each critical point

Check critical point

(\frac{5}{4}, - \frac{3}{2}) :

Minimum

Minimum (\frac{5}{4}, - \frac{3}{2})

f (x) = \frac{( - 3 x ^{3} + 8 x ^{2} - 5 x - 62 )}{x + 2}

f (x) = - 0.5 x^{2} + x

y = - 4 x^{3} + x

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

y = x 4 - x

What is the extreme f(x,y)=2x^2-5x+6y+2y^2 ?
The extreme f(x,y)=2x^2-5x+6y+2y^2 is Minimum(5/4 ,-3/2)