What is the extreme f(x)=2y^2+2xy+x^2-16x-20y ?

The extreme f(x)=2y^2+2xy+x^2-16x-20y is Minimum(6,2)

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extreme f(x)=2y^2+2xy+x^2-16x-20y

extreme points

f (x) = 2 y^{2} + 2 x y + x^{2} - 16 x - 20 y

Minimum (6, 2)

Solution steps

Find the critical points:

(6, 2)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(6, 2) :

Minimum

Minimum (6, 2)

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

f (x) = 4 x^{3} + 9 x^{2}

f (x, y) = x^{2} \cdot y^{2} + x^{2} - x \cdot y^{2} - 5 \cdot x - 7

\frac{3 x}{x - 2}

f (x, y) = 8 x y + \frac{1}{x} + \frac{1}{y}

What is the extreme f(x)=2y^2+2xy+x^2-16x-20y ?
The extreme f(x)=2y^2+2xy+x^2-16x-20y is Minimum(6,2)