What is the extreme f(x,y)= 1/2 x^2+2y^2-4x-16y ?

The extreme f(x,y)= 1/2 x^2+2y^2-4x-16y is Minimum(4,4)

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extreme f(x,y)= 1/2 x^2+2y^2-4x-16y

extreme points

f (x, y) = \frac{1}{2} x^{2} + 2 y^{2} - 4 x - 16 y

Minimum (4, 4)

Solution steps

Find the critical points:

(4, 4)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(4, 4) :

Minimum

Minimum (4, 4)

f (x) = \frac{x + 5}{x ^{2} - x - 30}

f (x) = - x^{2} + 9, - 4 \leq x \leq 3

f (x, y) = \frac{x ^{3} + 9}{x ^{2}}

f (x) = x^{2} + 10 x y + y^{2}

f (x) = 14 + 4 x - x^{2}

What is the extreme f(x,y)= 1/2 x^2+2y^2-4x-16y ?
The extreme f(x,y)= 1/2 x^2+2y^2-4x-16y is Minimum(4,4)