What is the extreme f(x,y)=8x^2+14xy+3y^2+10x-4 ?

The extreme f(x,y)=8x^2+14xy+3y^2+10x-4 is Saddle(3/5 ,-7/5)

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extreme f(x,y)=8x^2+14xy+3y^2+10x-4

extreme points

f (x, y) = 8 x^{2} + 14 x y + 3 y^{2} + 10 x - 4

Saddle (\frac{3}{5}, - \frac{7}{5})

Solution steps

Find the critical points:

(\frac{3}{5}, - \frac{7}{5})

\frac{\partial ^{2} f}{\partial x ^{2}} = 6

D (x, y) = - 100

Check the sign of at each critical point

Check critical point

(\frac{3}{5}, - \frac{7}{5}) :

Saddle

Saddle (\frac{3}{5}, - \frac{7}{5})

(x^{2} + 10) (9 - x^{2})

f (x, y) = x^{2} + y^{2} - 2 x + 6 y + 10

f (x) = \frac{1}{5} x^{5} - \frac{1}{4} x^{4} - 2 x^{3}

f (x) = x^{3} - 12 x^{2} + 48 x - 2

f (x) = \frac{1}{2} y^{4} - 4 x y + 2 x^{4}

What is the extreme f(x,y)=8x^2+14xy+3y^2+10x-4 ?
The extreme f(x,y)=8x^2+14xy+3y^2+10x-4 is Saddle(3/5 ,-7/5)