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

The extreme P(x,y)=3x^2y+4xy^2-y^4 is Maximum(2/3 ,-1)

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extreme P(x,y)=3x^2y+4xy^2-y^4

extreme points

P (x, y) = 3 x^{2} y + 4 x y^{2} - y^{4}

Maximum (\frac{2}{3}, - 1)

Solution steps

Find the critical points:

(\frac{2}{3}, - 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 8 x - 12 y^{2}

D (x, y) = - 72 y^{3} - 64 y^{2} - 48 x y - 36 x^{2}

Check the sign of at each critical point

Check critical point

(\frac{2}{3}, - 1) :

Maximum

Maximum (\frac{2}{3}, - 1)

f (x) = 3 x^{5} - 20 x^{3} + 16

\frac{x ^{3}}{x - 2}

f (x) = - x^{2} + 4, - 1 \leq x \leq 2

f (x) = 20 x^{3} + 9 x^{2} - 24 x + 12, 0 \leq x \leq 1

5 x^{3} - 30 x^{2} + 45 x

What is the extreme P(x,y)=3x^2y+4xy^2-y^4 ?
The extreme P(x,y)=3x^2y+4xy^2-y^4 is Maximum(2/3 ,-1)