What is the extreme f(x,y)=48xy-32x^3-24y^2 ?

The extreme f(x,y)=48xy-32x^3-24y^2 is Saddle(0,0),Maximum(1/2 , 1/2)

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extreme f(x,y)=48xy-32x^3-24y^2

extreme points

f (x, y) = 48 x y - 32 x^{3} - 24 y^{2}

Saddle (0, 0), Maximum (\frac{1}{2}, \frac{1}{2})

Solution steps

Find the critical points:

(0, 0), (\frac{1}{2}, \frac{1}{2})

\frac{\partial ^{2} f}{\partial x ^{2}} = - 48

D (x, y) = 9216 x - 2304

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

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

Maximum

Saddle (0, 0), Maximum (\frac{1}{2}, \frac{1}{2})

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

f (x) = 2 (3 - x) [- 1.2]

y = 6 x - x^{2}

f (x) = \frac{- 4 x + 92}{( x - 119 ) ^{7}}

f (x) = x^{\frac{1}{x}}

What is the extreme f(x,y)=48xy-32x^3-24y^2 ?
The extreme f(x,y)=48xy-32x^3-24y^2 is Saddle(0,0),Maximum(1/2 , 1/2)