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

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

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extreme f(xy)=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)

g (x) = 2 x^{3} + 3 x^{2} + 12 x - 4

f (x) = x^{3} - 10 x^{2} + 33 x - 36

x^{x}

x^{4} - 6 x^{3}

f (x) = x^{\frac{2}{3}} (x^{2} - 4)

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