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

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

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

extreme points

f (x, y) = 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)

f (x, y) = \frac{1}{x} - \frac{64}{y} + x y

f (x) = x 5 - x

f (x) = cos (3 x) - 2, (0, 2 π)

f (x) = \frac{1}{x + 1}

f (x) = - 3 sin (2 x) - 4

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