What is the extreme f(x,y)=8x-x^2-4y-y^2 ?

The extreme f(x,y)=8x-x^2-4y-y^2 is Maximum(4,-2)

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extreme f(x,y)=8x-x^2-4y-y^2

extreme points

f (x, y) = 8 x - x^{2} - 4 y - y^{2}

Maximum (4, - 2)

Solution steps

Find the critical points:

(4, - 2)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(4, - 2) :

Maximum

Maximum (4, - 2)

f (x) = x^{2} - 4 x - 6 [0.2]

f (x) = - 64 x^{3} + 12 x + 9

f (x, y) = - 3 x^{2} - 2 y^{2} + 24 x + 20 y + 600

f (x) = 5 (\frac{1}{x - 4} - \frac{1}{x + 2})

\frac{x ^{3} - 4 x ^{2}}{x ^{2} + 1}

What is the extreme f(x,y)=8x-x^2-4y-y^2 ?
The extreme f(x,y)=8x-x^2-4y-y^2 is Maximum(4,-2)