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

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

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extreme f(xy)=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) = 2 x^{3} + 3 x^{2} - 36 x + 17

f (x) = x^{3} + 2 x^{2} - 7 x [0.2]

T (x, y) = 26 - 9 x^{2} - 9 y^{2}

y = 6 x^{2} - 30 x + 9

f (x) = - 15 - 7 x - x^{2}

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