What is the extreme f(x,y)=2x^5+x^3y^2+6xy^4 ?

The extreme f(x,y)=2x^5+x^3y^2+6xy^4 is None

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extreme f(x,y)=2x^5+x^3y^2+6xy^4

extreme points

f (x, y) = 2 x^{5} + x^{3} y^{2} + 6 x y^{4}

None

Solution steps

Find the critical points:

None

2 x^{5} + x^{3} y^{2} + 6 x y^{4}

has no critical points, therefore there are no extreme points

None

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

f (x, y) = x^{2} y^{3} - (x - 2 y)^{2} + 6 x^{4} y = 3 x y

f (x) = 2 x^{4} - 20 x^{2} + 18

- x^{2} + 4 x + 6

f (x, y) = x^{2} y - 15 x y^{2} + 12 y

What is the extreme f(x,y)=2x^5+x^3y^2+6xy^4 ?
The extreme f(x,y)=2x^5+x^3y^2+6xy^4 is None