What is the extreme f(x,y)=5-2x+4y-x^2-4y^2 ?

The extreme f(x,y)=5-2x+4y-x^2-4y^2 is Maximum(-1, 1/2)

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extreme f(x,y)=5-2x+4y-x^2-4y^2

extreme points

f (x, y) = 5 - 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) = x - 2 x

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

f (x, y) = 2 x^{2} + y^{2} + 3 x y - 3 y - 5 x + 8

4 x^{3} - 3 x^{2} - 6 x + 3

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

What is the extreme f(x,y)=5-2x+4y-x^2-4y^2 ?
The extreme f(x,y)=5-2x+4y-x^2-4y^2 is Maximum(-1, 1/2)