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

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

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

extreme points

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

Maximum (1, 2)

Solution steps

Find the critical points:

(1, 2)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(1, 2) :

Maximum

Maximum (1, 2)

f (x) = 4 x^{3} - 36 x^{2} + 80 x

2 x - 4

f (x) = 4 x^{3} - 36 x^{2} + 81 x

y = (x - 1)^{3} + 3

f (x, y) = x 2 y

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