What is the extreme f(x,y)=7+2x+4y-x^2-4y^2 ?

The extreme f(x,y)=7+2x+4y-x^2-4y^2 is Maximum(1, 1/2)

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extreme f(x,y)=7+2x+4y-x^2-4y^2

extreme points

f (x, y) = 7 + 2 x + 4 y - x^{2} - 4 y^{2}

Maximum (1, \frac{1}{2})

Solution steps

Find the critical points:

(1, \frac{1}{2})

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

D (x, y) = 16

Check the sign of at each critical point

Check critical point

(1, \frac{1}{2}) :

Maximum

Maximum (1, \frac{1}{2})

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

f (x) = \frac{3 - x}{x + 2}

f (x) = 0.07 x^{3} - 1.6 x^{2} + 7.08 x + 5.48

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

What is the extreme f(x,y)=7+2x+4y-x^2-4y^2 ?
The extreme f(x,y)=7+2x+4y-x^2-4y^2 is Maximum(1, 1/2)