Menu Close

f-R-2-R-f-x-y-x-2-y-2-y-0-f-x-y-x-2-y-2-y-0-find-x-y-for-min-f-x-y-x-y-for-f-x-y-1-




Question Number 2524 by 123456 last updated on 21/Nov/15
f:R^2 →R  f(x,y)=x^2 −y^2           y≥0  f(x,y)=x^2 +y^2           y≤0  find  (x,y) for min f(x,y)  (x,y) for f(x,y)=1
$${f}:\mathbb{R}^{\mathrm{2}} \rightarrow\mathbb{R} \\ $$$${f}\left({x},{y}\right)={x}^{\mathrm{2}} −{y}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:{y}\geqslant\mathrm{0} \\ $$$${f}\left({x},{y}\right)={x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:{y}\leqslant\mathrm{0} \\ $$$$\mathrm{find} \\ $$$$\left({x},{y}\right)\:\mathrm{for}\:\mathrm{min}\:{f}\left({x},{y}\right) \\ $$$$\left({x},{y}\right)\:\mathrm{for}\:{f}\left({x},{y}\right)=\mathrm{1} \\ $$$$ \\ $$
Answered by prakash jain last updated on 22/Nov/15
f(x,y)=1  x^2 −y^2 =1⇒x^2 =1+y^2 ⇒x=±(√(1+y^2 )) ,y>0  x^2 +y^2 =1⇒x^2 =1−y^2 ⇒x=±(√(1−y^2 )), −1≤y≤0  min f(x,y)  There is no minimum value for f(x,y) since y can be  made arbitarilary large.
$${f}\left({x},{y}\right)=\mathrm{1} \\ $$$${x}^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{1}\Rightarrow{x}^{\mathrm{2}} =\mathrm{1}+{y}^{\mathrm{2}} \Rightarrow{x}=\pm\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }\:,{y}>\mathrm{0} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1}\Rightarrow{x}^{\mathrm{2}} =\mathrm{1}−{y}^{\mathrm{2}} \Rightarrow{x}=\pm\sqrt{\mathrm{1}−{y}^{\mathrm{2}} },\:−\mathrm{1}\leqslant{y}\leqslant\mathrm{0} \\ $$$$\mathrm{min}\:{f}\left({x},{y}\right) \\ $$$$\mathrm{There}\:\mathrm{is}\:\mathrm{no}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{for}\:{f}\left({x},{y}\right)\:\mathrm{since}\:{y}\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{made}\:\mathrm{arbitarilary}\:\mathrm{large}. \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *