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 Saddle(-2/3 ,1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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}

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

Solution steps

Find the critical points:

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

\frac{\partial ^{2} f}{\partial x ^{2}} = 8 x - \frac{20}{y ^{6}}

D (x, y) = - 64 y^{2} - 48 x y - \frac{120}{y ^{5}} - 36 x^{2}

Check the sign of at each critical point

Check critical point

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

Saddle

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

f (x) = (x + 10) (\frac{60}{x} + 6)

f (x) = \frac{1}{2} x^{4} - 4 x^{2} + 5

f (x) = \frac{1}{2} x^{4} - 4 x^{2} + 3

f (x) = - 3 x^{2} + 180 x + 190, 0 \leq x \leq 40

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

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 Saddle(-2/3 ,1)