What is the extreme f(x,y)=15x^2-2x^3+3y^2+6xy ?

The extreme f(x,y)=15x^2-2x^3+3y^2+6xy is Minimum(0,0),Saddle(4,-4)

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extreme f(x,y)=15x^2-2x^3+3y^2+6xy

extreme points

f (x, y) = 15 x^{2} - 2 x^{3} + 3 y^{2} + 6 x y

Minimum (0, 0), Saddle (4, - 4)

Solution steps

Find the critical points:

(0, 0), (4, - 4)

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

D (x, y) = - 72 x + 144

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Check critical point

(4, - 4) :

Saddle

Minimum (0, 0), Saddle (4, - 4)

4 x^{\frac{3}{4}} - x, 0 \leq x \leq 256

\frac{2 x + 1}{x ^{2} + 12 x + 48}

f (x) = x^{3} - 2 x^{2} + 3, (2, 7)

2 cos (x) - 2

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

What is the extreme f(x,y)=15x^2-2x^3+3y^2+6xy ?
The extreme f(x,y)=15x^2-2x^3+3y^2+6xy is Minimum(0,0),Saddle(4,-4)