Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 127088 by MathSh last updated on 26/Dec/20

{ ((x+y=k)),((x−3y=6)) :} (x^2 −y^2 )_(min) =?

$$\begin{cases}{{x}+{y}={k}}\\{{x}−\mathrm{3}{y}=\mathrm{6}}\end{cases} \\ $$$$\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)_{{min}} =? \\ $$

Commented by hknkrc46 last updated on 26/Dec/20

✓ x − 3y = (x + y) − 4y = k − 4y = 6 ⇒ 4y = k − 6 ⇒ y = ((k − 6)/4) ✓ x + y = x + ((k − 6)/4) = k ⇒ x = k − ((k − 6)/4) = ((3k + 6)/4) ✓ x^2 − y^2 = (((3k + 6)/4))^2 − (((k − 6)/4))^2 = ((k^2 + 6k)/2) = (((k + 3)^2 − 9)/2) (min : k = −3) = − (9/2)

$$\checkmark\:\boldsymbol{{x}}\:−\:\mathrm{3}\boldsymbol{{y}}\:=\:\left(\boldsymbol{{x}}\:+\:\boldsymbol{{y}}\right)\:−\:\mathrm{4}\boldsymbol{{y}}\:=\:\boldsymbol{{k}}\:−\:\mathrm{4}\boldsymbol{{y}}\:=\:\mathrm{6} \\ $$$$\Rightarrow\:\mathrm{4}\boldsymbol{{y}}\:=\:\boldsymbol{{k}}\:−\:\mathrm{6}\:\Rightarrow\:\boldsymbol{{y}}\:=\:\frac{\boldsymbol{{k}}\:−\:\mathrm{6}}{\mathrm{4}} \\ $$$$\checkmark\:\boldsymbol{{x}}\:+\:\boldsymbol{{y}}\:=\:\boldsymbol{{x}}\:+\:\frac{\boldsymbol{{k}}\:−\:\mathrm{6}}{\mathrm{4}}\:=\:\boldsymbol{{k}} \\ $$$$\Rightarrow\:\boldsymbol{{x}}\:=\:\boldsymbol{{k}}\:−\:\frac{\boldsymbol{{k}}\:−\:\mathrm{6}}{\mathrm{4}}\:=\:\frac{\mathrm{3}\boldsymbol{{k}}\:+\:\mathrm{6}}{\mathrm{4}} \\ $$$$\checkmark\:\boldsymbol{{x}}^{\mathrm{2}} \:−\:\boldsymbol{{y}}^{\mathrm{2}} \:=\:\left(\frac{\mathrm{3}\boldsymbol{{k}}\:+\:\mathrm{6}}{\mathrm{4}}\right)^{\mathrm{2}} −\:\left(\frac{\boldsymbol{{k}}\:−\:\mathrm{6}}{\mathrm{4}}\right)^{\mathrm{2}} =\:\frac{\boldsymbol{{k}}^{\mathrm{2}} \:+\:\mathrm{6}\boldsymbol{{k}}}{\mathrm{2}} \\ $$$$=\:\frac{\left(\boldsymbol{{k}}\:+\:\mathrm{3}\right)^{\mathrm{2}} \:−\:\mathrm{9}}{\mathrm{2}}\:\left(\boldsymbol{{min}}\::\:\boldsymbol{{k}}\:=\:−\mathrm{3}\right)\:=\:−\:\frac{\mathrm{9}}{\mathrm{2}} \\ $$

Commented by MathSh last updated on 27/Dec/20

Thanks sir

$${Thanks}\:{sir} \\ $$

Answered by Olaf last updated on 26/Dec/20

y = ((k−6)/4), x = ((3(k+2))/4) x^2 −y^2 = (9/(16))(k+2)^2 −(1/(16))(k−6)^2 x^2 −y^2 = (9/(16))(k^2 +4k+4)−(1/(16))(k^2 −12k+36) x^2 −y^2 = (1/2)k^2 +3k x^2 −y^2 = (k/2)(k+6) If k = −3, (x^2 −y^2 )_(min) = −(9/2) x = −(3/4) y = −(9/4)

$$ \\ $$$${y}\:=\:\frac{{k}−\mathrm{6}}{\mathrm{4}},\:{x}\:=\:\frac{\mathrm{3}\left({k}+\mathrm{2}\right)}{\mathrm{4}} \\ $$$${x}^{\mathrm{2}} −{y}^{\mathrm{2}} \:=\:\frac{\mathrm{9}}{\mathrm{16}}\left({k}+\mathrm{2}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{16}}\left({k}−\mathrm{6}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −{y}^{\mathrm{2}} \:=\:\frac{\mathrm{9}}{\mathrm{16}}\left({k}^{\mathrm{2}} +\mathrm{4}{k}+\mathrm{4}\right)−\frac{\mathrm{1}}{\mathrm{16}}\left({k}^{\mathrm{2}} −\mathrm{12}{k}+\mathrm{36}\right) \\ $$$${x}^{\mathrm{2}} −{y}^{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}{k}^{\mathrm{2}} +\mathrm{3}{k} \\ $$$${x}^{\mathrm{2}} −{y}^{\mathrm{2}} \:=\:\frac{{k}}{\mathrm{2}}\left({k}+\mathrm{6}\right) \\ $$$$\mathrm{If}\:{k}\:=\:−\mathrm{3},\:\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)_{{min}} \:=\:−\frac{\mathrm{9}}{\mathrm{2}} \\ $$$${x}\:\:=\:−\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${y}\:=\:−\frac{\mathrm{9}}{\mathrm{4}} \\ $$

Commented by MathSh last updated on 27/Dec/20

Thanks sir

$${Thanks}\:{sir} \\ $$

Answered by Olaf last updated on 26/Dec/20

sorry sir ! I read x^2 +y^2 instead of x^2 −y^2 . I′m tired !

$${sorry}\:{sir}\:!\:{I}\:{read}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:{instead}\:{of} \\ $$$${x}^{\mathrm{2}} −{y}^{\mathrm{2}} .\:{I}'{m}\:{tired}\:! \\ $$

Commented by MathSh last updated on 27/Dec/20

Yes, thanks sir

$${Yes},\:{thanks}\:{sir} \\ $$

Answered by mathmax by abdo last updated on 26/Dec/20

{ ((x+y=k)),((x−3y=6 ⇒ { ((4y=k−6 ⇒ { ((y=((k−6)/4))),((x=k−((k−6)/4))) :})),((x=k−y)) :})) :} ⇒ { ((x=((3k+6)/4))),((y=((k−6)/4) ⇒ x^2 −y^2 =(1/(16)){(3k+6)^2 −(k−6)^2 }=f(k) ⇒)) :} f(k)=(1/(16))(9k^2 +36k +36−k^2 +12k−36} =(1/(16))(8k^2 +48k) =(k^2 /2) +3k ⇒f^′ (k)=k+3 and f^′ (k)=0 ⇒k=−3 k −∞ −3 +∞ f^′ − 0 + f decr f(−3) incr ⇒ minf(k)=f(−3) =(9/2)−9 =−(9/2) =min(x^2 −y^2 )

$$\begin{cases}{\mathrm{x}+\mathrm{y}=\mathrm{k}}\\{\mathrm{x}−\mathrm{3y}=\mathrm{6}\:\:\:\:\:\:\Rightarrow\:\begin{cases}{\mathrm{4y}=\mathrm{k}−\mathrm{6}\:\:\Rightarrow\begin{cases}{\mathrm{y}=\frac{\mathrm{k}−\mathrm{6}}{\mathrm{4}}}\\{\mathrm{x}=\mathrm{k}−\frac{\mathrm{k}−\mathrm{6}}{\mathrm{4}}}\end{cases}}\\{\mathrm{x}=\mathrm{k}−\mathrm{y}}\end{cases}}\end{cases} \\ $$$$\Rightarrow\begin{cases}{\mathrm{x}=\frac{\mathrm{3k}+\mathrm{6}}{\mathrm{4}}}\\{\mathrm{y}=\frac{\mathrm{k}−\mathrm{6}}{\mathrm{4}}\:\Rightarrow\:\:\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \:=\frac{\mathrm{1}}{\mathrm{16}}\left\{\left(\mathrm{3k}+\mathrm{6}\right)^{\mathrm{2}} −\left(\mathrm{k}−\mathrm{6}\right)^{\mathrm{2}} \right\}=\mathrm{f}\left(\mathrm{k}\right)\:\Rightarrow}\end{cases} \\ $$$$\mathrm{f}\left(\mathrm{k}\right)=\frac{\mathrm{1}}{\mathrm{16}}\left(\mathrm{9k}^{\mathrm{2}} +\mathrm{36k}\:+\mathrm{36}−\mathrm{k}^{\mathrm{2}} +\mathrm{12k}−\mathrm{36}\right\}\:=\frac{\mathrm{1}}{\mathrm{16}}\left(\mathrm{8k}^{\mathrm{2}} +\mathrm{48k}\right) \\ $$$$=\frac{\mathrm{k}^{\mathrm{2}} }{\mathrm{2}}\:+\mathrm{3k}\:\:\Rightarrow\mathrm{f}^{'} \left(\mathrm{k}\right)=\mathrm{k}+\mathrm{3}\:\:\mathrm{and}\:\mathrm{f}^{'} \left(\mathrm{k}\right)=\mathrm{0}\:\Rightarrow\mathrm{k}=−\mathrm{3} \\ $$$$\mathrm{k}\:\:\:\:\:\:\:\:\:\:\:\:−\infty\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\infty \\ $$$$\mathrm{f}^{'} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:+ \\ $$$$\mathrm{f}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{decr}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{f}\left(−\mathrm{3}\right)\:\:\:\mathrm{incr}\:\:\Rightarrow \\ $$$$\mathrm{minf}\left(\mathrm{k}\right)=\mathrm{f}\left(−\mathrm{3}\right)\:=\frac{\mathrm{9}}{\mathrm{2}}−\mathrm{9}\:\:=−\frac{\mathrm{9}}{\mathrm{2}}\:=\mathrm{min}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \right) \\ $$

Commented by MathSh last updated on 27/Dec/20

Thanks sir

$${Thanks}\:{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com