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

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

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

extreme points

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

Minimum (2, 1)

Solution steps

Find the critical points:

(2, 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 2 + \frac{4}{y ^{3}}

D (x, y) = (\frac{4}{y ^{3}} + 2) (\frac{32}{x ^{3}} + 2)

Check the sign of at each critical point

Check critical point

(2, 1) :

Minimum

Minimum (2, 1)

f (x) = - 5 x^{4} - 5 x^{3} - 2

f (x) = - x^{3} - 6 x^{2} - 1

f (x) = \frac{x ^{2} - 4}{x ^{2} + 4}, - 5 \leq x \leq 5

f (x) = 7 x^{2} - 56 x + 84

y = - 4 x^{2} - 5 x + 1

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