What is the extreme f(x,y)=2x^2+2y^2-16x+12y ?

The extreme f(x,y)=2x^2+2y^2-16x+12y is Minimum(4,-3)

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extreme f(x,y)=2x^2+2y^2-16x+12y

extreme points

f (x, y) = 2 x^{2} + 2 y^{2} - 16 x + 12 y

Minimum (4, - 3)

Solution steps

Find the critical points:

(4, - 3)

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

D (x, y) = 16

Check the sign of at each critical point

Check critical point

(4, - 3) :

Minimum

Minimum (4, - 3)

f (x) = x^{3} - 15 x^{2} + 63 x - 15

f (x, y) = \frac{8460}{x} + \frac{8460}{y} + 5 x y

f (x) = (85 - 0.05 x) - (600 + 35 x)

f (x) = 4 x^{2} - 24 x + 400

f (x, y) = - 5 x^{2} - 6 y^{2} + 5 x - 12 y + 5

What is the extreme f(x,y)=2x^2+2y^2-16x+12y ?
The extreme f(x,y)=2x^2+2y^2-16x+12y is Minimum(4,-3)