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

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

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

extreme points

f (x, y) = 4 - 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) = 2 x^{\frac{5}{3}} - 5 x^{\frac{4}{3}}

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

y = x (x + 4)^{3}

f (x) = \frac{4 x}{x ^{2} - 49}

f (x) = x^{3} - 6 x^{2} + 8, - 1 \leq x \leq 6

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