If-x-y-gt-0-then-the-minimum-value-of-2x-2-2-x-2x-2y-2-2-y-2y-2-is-equal-to-

Question Number 25001 by Tinkutara last updated on 30/Nov/17

If x, y > 0, then the minimum value of

2 x^{2} + \frac{2}{x} - 2 x + 2 y^{2} + \frac{2}{y} - 2 y + 2 is

equal to

Commented by prakash jain last updated on 01/Dec/17

u (x, y) = f (x) + f (y) + 2

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

f^{'} (x) = 4 x - \frac{2}{x^{2}} - 2 = 0

2 x^{3} - x^{2} - 1 = 0

2 x^{3} - 2 x^{2} + x^{2} - 1 = 0

2 x^{2} (x - 1) + (x + 1) (x - 1) = 0

(x - 1) (2 x^{2} + x + 1) = 0

\Rightarrow x = 1

m i n a t x = y = 1

u (1, 1) = 6

Commented by Tinkutara last updated on 01/Dec/17

Thank you Sir!

Commented by Tinkutara last updated on 01/Dec/17

M y m e t h o d :

E x p r e s s i o n i s r e w r i t t e n a s

{(x - 1)}^{2} + {(y - 1)}^{2} + (x^{2} + \frac{2}{x}) + (y^{2} + \frac{2}{y})

⩾ 0 + 0 + 3 + 3 = 6

Commented by prakash jain last updated on 01/Dec/17

Try your logic for the below

and compare with actual minimum

{(x - 1)}^{2} + {(y - 1)}^{2} + x^{2} + y^{2}

Commented by prakash jain last updated on 01/Dec/17

in general

f (x) = u (x) + v (x)

Minimizing u (x) does not guaratee

that f (x) is also minimum .