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

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

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

extreme points

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

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

Solution steps

Find the critical points:

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

\frac{\partial ^{2} f}{\partial x ^{2}} = - 4

D (x, y) = 4

Check the sign of at each critical point

Check critical point

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

Maximum

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

f (x) = - x^{3} + 3 x^{2} + 45 x - 42

f (x) = (x^{2} + 12) (1 - x^{2})

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

f (x) = \frac{- 3}{x ^{2} - 9}

f (x, y) = x y + \frac{9}{x} + \frac{3}{y}

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