What is the extreme f(x)=1-2x+4y-x^2-4y^2 ?

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

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extreme f(x)=1-2x+4y-x^2-4y^2

extreme points

f (x) = 1 - 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})

x^{2} = - y + 4

f (x, y) = - 4 x^{2} + 8 x - 9 y^{2} - 18 y - 12

f (x) = \frac{- x ^{3}}{3} - 2 x^{2} + 5 x - 2

f (x) = \frac{x}{4} + y - \frac{7}{4}

f (x, y) = \frac{1}{12} 144 - 16 x^{2} - 9 y^{2}

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