What is the extreme f(x)=8-2x+4y-x^2-4y^2 ?

The extreme f(x)=8-2x+4y-x^2-4y^2 is Maximum(-1, 1/2)

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extreme f(x)=8-2x+4y-x^2-4y^2

extreme points

f (x) = 8 - 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, y) = x^{2} + 2 x y + 2 y^{2} - 6 x + 10 y + 2

f (x) = \frac{x ^{2}}{1 - x}

f (x) = 4 - x^{2}, - 1 \leq x \leq 2

f (x) = \frac{1}{1 + ∣ x ∣} + \frac{1}{1 + ∣ x - 2 ∣}

f (x) = \frac{3 x}{x ^{2} + 9}

What is the extreme f(x)=8-2x+4y-x^2-4y^2 ?
The extreme f(x)=8-2x+4y-x^2-4y^2 is Maximum(-1, 1/2)