What is the extreme-x^2-4y^2+8x-8y-16 ?

The extreme-x^2-4y^2+8x-8y-16 is Maximum(4,-1)

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extreme-x^2-4y^2+8x-8y-16

extreme points

- x^{2} - 4 y^{2} + 8 x - 8 y - 16

Maximum (4, - 1)

Solution steps

Find the critical points:

(4, - 1)

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

D (x, y) = 16

Check the sign of at each critical point

Check critical point

(4, - 1) :

Maximum

Maximum (4, - 1)

f (x) = (x - 1)^{2} + y^{2} = 25

f (x) = 2 x^{2} ln (x) - 5 x^{2}

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

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

U (X, Y) = X^{0.4} Y^{0.6}

What is the extreme-x^2-4y^2+8x-8y-16 ?
The extreme-x^2-4y^2+8x-8y-16 is Maximum(4,-1)