What is the extreme f(x,y)=8+76xy+38x^2+240y+(y^4)/4 ?

The extreme f(x,y)=8+76xy+38x^2+240y+(y^4)/4 is Saddle(-4,4),Minimum(-6,6),Minimum(10,-10)

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extreme f(x,y)=8+76xy+38x^2+240y+(y^4)/4

extreme points

f (x, y) = 8 + 76 x y + 38 x^{2} + 240 y + \frac{y ^{4}}{4}

Saddle (- 4, 4), Minimum (- 6, 6), Minimum (10, - 10)

Solution steps

Find the critical points:

(- 4, 4), (- 6, 6), (10, - 10)

\frac{\partial ^{2} f}{\partial x ^{2}} = 3 y^{2}

D (x, y) = 228 y^{2} - 5776

Check the sign of at each critical point

Check critical point

(- 4, 4) :

Saddle

Check critical point

(- 6, 6) :

Minimum

Check critical point

(10, - 10) :

Minimum

Saddle (- 4, 4), Minimum (- 6, 6), Minimum (10, - 10)

y = \frac{x ^{2}}{ln ( x )}

f (x) = x^{3} - 6 x^{2} + 9 + 9, - 1 \leq x \leq 9

f (x, y) = 16 - x^{2} - 16 y^{2}

f (x) = x^{3} + y^{3} + 9 x^{2} - 3 y^{2} - 8

f (x, y) = 6 x - x^{2} + 2 y^{4} - 16 y^{2} + 5

What is the extreme f(x,y)=8+76xy+38x^2+240y+(y^4)/4 ?
The extreme f(x,y)=8+76xy+38x^2+240y+(y^4)/4 is Saddle(-4,4),Minimum(-6,6),Minimum(10,-10)