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

If x, y > 0, then the minimum value of 2x^2 + (2/x) − 2x + 2y^2 + (2/y) − 2y + 2 is equal to

$$\mathrm{If}\:{x},\:{y}\:>\:\mathrm{0},\:\mathrm{then}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of} \\ $$ $$\mathrm{2}{x}^{\mathrm{2}} \:+\:\frac{\mathrm{2}}{{x}}\:−\:\mathrm{2}{x}\:+\:\mathrm{2}{y}^{\mathrm{2}} \:+\:\frac{\mathrm{2}}{{y}}\:−\:\mathrm{2}{y}\:+\:\mathrm{2}\:\mathrm{is} \\ $$ $$\mathrm{equal}\:\mathrm{to} \\ $$

Commented byprakash jain last updated on 01/Dec/17

u(x,y)=f(x)+f(y)+2 f(x)=2x^2 +(2/x)−2x f′(x)=4x−(2/x^2 )−2=0 2x^3 −x^2 −1=0 2x^3 −2x^2 +x^2 −1=0 2x^2 (x−1)+(x+1)(x−1)=0 (x−1)(2x^2 +x+1)=0 ⇒x=1 min at x=y=1 u(1,1)=6

$${u}\left({x},{y}\right)={f}\left({x}\right)+{f}\left({y}\right)+\mathrm{2} \\ $$ $${f}\left({x}\right)=\mathrm{2}{x}^{\mathrm{2}} +\frac{\mathrm{2}}{{x}}−\mathrm{2}{x} \\ $$ $${f}'\left({x}\right)=\mathrm{4}{x}−\frac{\mathrm{2}}{{x}^{\mathrm{2}} }−\mathrm{2}=\mathrm{0} \\ $$ $$\mathrm{2}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$ $$\mathrm{2}{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$ $$\mathrm{2}{x}^{\mathrm{2}} \left({x}−\mathrm{1}\right)+\left({x}+\mathrm{1}\right)\left({x}−\mathrm{1}\right)=\mathrm{0} \\ $$ $$\left({x}−\mathrm{1}\right)\left(\mathrm{2}{x}^{\mathrm{2}} +{x}+\mathrm{1}\right)=\mathrm{0} \\ $$ $$\Rightarrow{x}=\mathrm{1} \\ $$ $${min}\:{at}\:{x}={y}=\mathrm{1} \\ $$ $${u}\left(\mathrm{1},\mathrm{1}\right)=\mathrm{6} \\ $$

Commented byTinkutara last updated on 01/Dec/17

Thank you Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir}! \\ $$

Commented byTinkutara last updated on 01/Dec/17

My method: Expression is rewritten as (x−1)^2 +(y−1)^2 +(x^2 +(2/x))+(y^2 +(2/y)) ≥0+0+3+3=6

$${My}\:{method}: \\ $$ $${Expression}\:{is}\:{rewritten}\:{as} \\ $$ $$\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} +\left({x}^{\mathrm{2}} +\frac{\mathrm{2}}{{x}}\right)+\left({y}^{\mathrm{2}} +\frac{\mathrm{2}}{{y}}\right) \\ $$ $$\geqslant\mathrm{0}+\mathrm{0}+\mathrm{3}+\mathrm{3}=\mathrm{6} \\ $$

Commented byprakash 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

$$\mathrm{Try}\:\mathrm{your}\:\mathrm{logic}\:\mathrm{for}\:\mathrm{the}\:\mathrm{below} \\ $$ $$\mathrm{and}\:\mathrm{compare}\:\mathrm{with}\:\mathrm{actual}\:\mathrm{minimum} \\ $$ $$\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} +{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \\ $$

Commented byprakash 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.

$$\mathrm{in}\:\mathrm{general} \\ $$ $${f}\left({x}\right)={u}\left({x}\right)+{v}\left({x}\right) \\ $$ $$\mathrm{Minimizing}\:{u}\left({x}\right)\:\mathrm{does}\:\mathrm{not}\:\mathrm{guaratee} \\ $$ $$\mathrm{that}\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{also}\:\mathrm{minimum}. \\ $$

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