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

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

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extreme f(x)=6x-4y-x^2-2y^2

extreme points

f (x) = 6 x - 4 y - x^{2} - 2 y^{2}

Maximum (3, - 1)

Solution steps

Find the critical points:

(3, - 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 4

D (x, y) = 8

Check the sign of at each critical point

Check critical point

(3, - 1) :

Maximum

Maximum (3, - 1)

f (x) = x^{4} - 6 x^{2} + 9

f (x) = - x^{2} + 8 x - 15

f (x) = x^{\frac{4}{5}} (9 - 4 x), 0 \leq x \leq 2

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

f (x) = 2 x^{3} - 13 x^{2} + 24 x - 28

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