What is the extreme 9-2x+4y-x^2-4y^2 ?

The extreme 9-2x+4y-x^2-4y^2 is Maximum(-1, 1/2)

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extreme 9-2x+4y-x^2-4y^2

extreme points

9 - 2 x + 4 y - x^{2} - 4 y^{2}

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

Solution steps

Find the critical points:

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

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

D (x, y) = 16

Check the sign of at each critical point

Check critical point

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

Maximum

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

f (x) = 6 x + 2 x y - 2 y - 2 x^{2} - y^{2}

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

f (x) = - 5 cos (x - 3 π)

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

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

What is the extreme 9-2x+4y-x^2-4y^2 ?
The extreme 9-2x+4y-x^2-4y^2 is Maximum(-1, 1/2)