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

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

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

extreme points

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

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) = x 81 - x^{2}

S (a, d) = a + (n - 1) d

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

y = 3 x + 2 cos (x), 0 \leq x \leq 2 π

f (x) = 2 x - \frac{108}{x}

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