What is the extreme f(x,y)=2x^2-8x+y^2+16y+100 ?

The extreme f(x,y)=2x^2-8x+y^2+16y+100 is Minimum(2,-8)

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extreme f(x,y)=2x^2-8x+y^2+16y+100

extreme points

f (x, y) = 2 x^{2} - 8 x + y^{2} + 16 y + 100

Minimum (2, - 8)

Solution steps

Find the critical points:

(2, - 8)

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

D (x, y) = 8

Check the sign of at each critical point

Check critical point

(2, - 8) :

Minimum

Minimum (2, - 8)

f (x, y) = \frac{1}{25 - x ^{2} - y ^{2}}

f (x) = 1 - x^{2}

f (x, y) = 4 x^{2} y + 2 x y^{2} - 12 x y - 5

f (x) = x^{3} + \frac{48}{x}

f (x, y) = \frac{y ^{2}}{y + x ^{2}}

What is the extreme f(x,y)=2x^2-8x+y^2+16y+100 ?
The extreme f(x,y)=2x^2-8x+y^2+16y+100 is Minimum(2,-8)