What is the extreme f(x)=-2x^2-2y^2+20x+16y+4 ?

The extreme f(x)=-2x^2-2y^2+20x+16y+4 is Maximum(5,4)

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extreme f(x)=-2x^2-2y^2+20x+16y+4

extreme points

f (x) = - 2 x^{2} - 2 y^{2} + 20 x + 16 y + 4

Maximum (5, 4)

Solution steps

Find the critical points:

(5, 4)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 4

D (x, y) = 16

Check the sign of at each critical point

Check critical point

(5, 4) :

Maximum

Maximum (5, 4)

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

\frac{1}{3} x^{3} - \frac{1}{2} x^{2} - 2 x

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

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

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

What is the extreme f(x)=-2x^2-2y^2+20x+16y+4 ?
The extreme f(x)=-2x^2-2y^2+20x+16y+4 is Maximum(5,4)