What is the extreme f(x,y)=60x^{1/3}y^{2/3} ?

The extreme f(x,y)=60x^{1/3}y^{2/3} is None

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extreme f(x,y)=60x^{1/3}y^{2/3}

extreme points

f (x, y) = 60 x^{\frac{1}{3}} y^{\frac{2}{3}}

None

Solution steps

Find the critical points:

None

60 x^{\frac{1}{3}} y^{\frac{2}{3}}

has no critical points, therefore there are no extreme points

None

x^{2} (x^{2} - 2)

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

- 3 x^{3}

y = x^{5} - 5 x^{4}

f (x) = x^{3} + 3 x^{2}, - \frac{5}{2} \leq x \leq 1

What is the extreme f(x,y)=60x^{1/3}y^{2/3} ?
The extreme f(x,y)=60x^{1/3}y^{2/3} is None