What is the extreme f(x,y)=2x^2+3xy+4y^2-6x+7y ?

The extreme f(x,y)=2x^2+3xy+4y^2-6x+7y is Minimum(3,-2)

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extreme f(x,y)=2x^2+3xy+4y^2-6x+7y

extreme points

f (x, y) = 2 x^{2} + 3 x y + 4 y^{2} - 6 x + 7 y

Minimum (3, - 2)

Solution steps

Find the critical points:

(3, - 2)

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

D (x, y) = 23

Check the sign of at each critical point

Check critical point

(3, - 2) :

Minimum

Minimum (3, - 2)

f (x) = 3 x^{4} - 16 x^{3} + 18 x^{2}, - 1 \leq x \leq 4

x^{\frac{2}{3}} (6 - x)^{\frac{1}{3}}

f (x) = x^{8} e^{5 x}

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

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

What is the extreme f(x,y)=2x^2+3xy+4y^2-6x+7y ?
The extreme f(x,y)=2x^2+3xy+4y^2-6x+7y is Minimum(3,-2)