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

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

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

extreme points

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

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) = 48 π r^{2} + \frac{1804.8}{r}

f (x) = \frac{x + 1}{x ^{2} - 4}

f (x) = (x - 4) ln (x - 4)

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

f (x, y) = 5 x^{2} + x y^{2} + y^{2}

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