Question Number 103040 by bobhans last updated on 12/Jul/20
$${given}\:\mathrm{5}{x}+\mathrm{12}{y}\:=\:\mathrm{60} \\ $$$${min}\:{value}\:{of}\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} } \\ $$
Answered by bobhans last updated on 12/Jul/20
Commented by mathmax by abdo last updated on 12/Jul/20
$$\mathrm{error}\:\mathrm{sir}\:\:\frac{\partial\mathrm{f}}{\partial\lambda}=\mathrm{5x}\:+\mathrm{2y}−\mathrm{60}\:! \\ $$
Commented by bemath last updated on 12/Jul/20
$${the}\:{equation}\:\mathrm{5}{x}+\mathrm{12}{y}−\mathrm{60}=\mathrm{0}\:{sir} \\ $$$${it}\:{typo}\: \\ $$$$ \\ $$$$ \\ $$
Answered by ajfour last updated on 12/Jul/20
$${let}\:{x}={r}\mathrm{cos}\:\theta\:,\:{y}={r}\mathrm{sin}\:\theta \\ $$$$\Rightarrow\:\:\:{r}=\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} } \\ $$$$\:\:\:\mathrm{13}\left(\frac{\mathrm{5}{x}}{\mathrm{13}}+\frac{\mathrm{12}{y}}{\mathrm{13}}\right)=\mathrm{60} \\ $$$$\left\{{if}\:\:\:\mathrm{sin}\:\alpha=\frac{\mathrm{5}}{\mathrm{13}}\:,\:\:\mathrm{cos}\:\alpha=\frac{\mathrm{12}}{\mathrm{13}}\right. \\ $$$$\left.\Rightarrow\:\:\:\:\:\mathrm{tan}\:\alpha=\frac{\mathrm{5}}{\mathrm{12}}\:\right\} \\ $$$$\:\:\mathrm{13}{r}\left(\mathrm{sin}\:\alpha\mathrm{cos}\:\theta+\mathrm{cos}\:\alpha\mathrm{sin}\:\theta\right)=\mathrm{60} \\ $$$$\Rightarrow\:\:\mathrm{13}{r}\mathrm{sin}\:\left(\alpha+\theta\right)=\mathrm{60} \\ $$$$\Rightarrow\:\:\:\:{r}=\frac{\mathrm{60}}{\mathrm{13sin}\left(\alpha+\theta\right)} \\ $$$${and}\:\:{as}\:\:{r}>\mathrm{0}\:\:\:\:\Rightarrow\:\:{r}_{{min}} =\frac{\mathrm{60}}{\mathrm{13}} \\ $$$${when}\:\:\theta=\mathrm{2}{n}\pi+\frac{\pi}{\mathrm{2}}−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{5}}{\mathrm{12}}\:\:. \\ $$
Answered by maths mind last updated on 12/Jul/20
$${M}\left({x},{y}\right)\in\left({D}\right):\:\mathrm{5}{x}+\mathrm{12}{y}−\mathrm{60}=\mathrm{0} \\ $$$$\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }={d}\left({O},{M}\right) \\ $$$${min}\:{d}\left({O},{M}\right)={d}\left({O},{M}'\right) \\ $$$${M}'\:{ptojection}\:{of}\:{O}\:{over}\:\left({d}\right) \\ $$$${d}\left(\mathrm{0},{M}'\right)=\frac{\mid\mathrm{0}.\mathrm{5}+\mathrm{12}.\mathrm{0}−\mathrm{60}\mid}{\:\sqrt{\mathrm{5}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} }}=\frac{\mathrm{60}}{\mathrm{13}} \\ $$
Answered by mr W last updated on 12/Jul/20
$${say}\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }={D} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={D}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +\left(\mathrm{5}−\frac{\mathrm{5}{x}}{\mathrm{12}}\right)^{\mathrm{2}} ={D}^{\mathrm{2}} \\ $$$$\frac{\mathrm{169}}{\mathrm{144}}{x}^{\mathrm{2}} −\frac{\mathrm{50}}{\mathrm{12}}{x}+\mathrm{25}−{D}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Delta=\left(\frac{\mathrm{50}}{\mathrm{12}}\right)^{\mathrm{2}} −\mathrm{4}×\frac{\mathrm{169}}{\mathrm{144}}\left(\mathrm{25}−{D}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${D}^{\mathrm{2}} =\frac{\mathrm{3600}}{\mathrm{169}} \\ $$$$\Rightarrow{D}=\frac{\mathrm{60}}{\mathrm{13}} \\ $$
Answered by 1549442205 last updated on 13/Jul/20
$$\mathrm{Apply}\:\mathrm{Cauchy}−\mathrm{Shward}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{60}^{\mathrm{2}} =\left(\mathrm{5x}+\mathrm{12y}\right)^{\mathrm{2}} \leqslant\left(\mathrm{5}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} \right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \geqslant\frac{\mathrm{60}^{\mathrm{2}} }{\mathrm{5}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} }=\frac{\mathrm{60}^{\mathrm{2}} }{\mathrm{13}^{\mathrm{2}} }\Rightarrow\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }\geqslant\frac{\mathrm{60}}{\mathrm{13}} \\ $$$$\mathrm{the}\:\mathrm{equality}\:\mathrm{ocurrs}\:\mathrm{if}\:\mathrm{and}\:\mathrm{only}\:\mathrm{if} \\ $$$$\begin{cases}{\frac{\mathrm{x}}{\mathrm{5}}=\frac{\mathrm{y}}{\mathrm{12}}}\\{\mathrm{5x}+\mathrm{12y}=\mathrm{60}}\end{cases}\Leftrightarrow\begin{cases}{\mathrm{x}=\frac{\mathrm{300}}{\mathrm{169}}}\\{\mathrm{y}=\frac{\mathrm{720}}{\mathrm{169}}}\end{cases} \\ $$$$\boldsymbol{\mathrm{Thus}},\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{y}}^{\mathrm{2}} }\:\boldsymbol{\mathrm{has}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{smallest}}\:\boldsymbol{\mathrm{value}} \\ $$$$\boldsymbol{\mathrm{equal}}\:\frac{\mathrm{60}}{\mathrm{13}}\:\boldsymbol{\mathrm{when}}\:\left(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}}\right)=\left(\frac{\mathrm{300}}{\mathrm{169}};\frac{\mathrm{720}}{\mathrm{169}}\right) \\ $$