What is the extreme f(x,y,z)=-x^2-3y^2+4x-6y-5 ?

The extreme f(x,y,z)=-x^2-3y^2+4x-6y-5 is Maximum(2,-1)

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extreme f(x,y,z)=-x^2-3y^2+4x-6y-5

extreme points

f (x, y, z) = - x^{2} - 3 y^{2} + 4 x - 6 y - 5

Maximum (2, - 1)

Solution steps

Find the critical points:

(2, - 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 6

D (x, y) = 12

Check the sign of at each critical point

Check critical point

(2, - 1) :

Maximum

Maximum (2, - 1)

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

f (x) = 4 x + 3

f (z) = x^{2} - y^{2} - 2 x + 4 y + 6

f (x) = \frac{x ^{3}}{3} - x^{2} - 3 x - 1, 2 \leq x \leq 7

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

What is the extreme f(x,y,z)=-x^2-3y^2+4x-6y-5 ?
The extreme f(x,y,z)=-x^2-3y^2+4x-6y-5 is Maximum(2,-1)