Algebra Questions

Question Number 79190 by mr W last updated on 23/Jan/20

$${if}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{50}, \\$$$${find}\:{the}\:{minimum}\:{and}\:{maximum}\:{of} \\$$$$\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{8}\left({x}+{y}\right)+\mathrm{20} \\$$

Commented by jagoll last updated on 23/Jan/20

$$\Rightarrow\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{2xy}−\mathrm{8x}−\mathrm{8y}+\mathrm{20}\:= \\$$$$\mathrm{2xy}−\mathrm{8x}−\mathrm{8y}+\mathrm{70} \\$$$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x},\mathrm{y},\lambda\right)\:=\:\mathrm{2xy}−\mathrm{8x}−\mathrm{8y}+\mathrm{70}+\lambda\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{50}\right) \\$$

Commented by jagoll last updated on 23/Jan/20

$$\mathrm{we}\:\mathrm{use}\:\mathrm{differential} \\$$

Commented by jagoll last updated on 23/Jan/20

$$\mathrm{right}\:\mathrm{sir}? \\$$

Commented by jagoll last updated on 23/Jan/20

$$\frac{\mathrm{df}}{\mathrm{dx}}=\mathrm{2y}−\mathrm{8}+\lambda\left(\mathrm{2x}\right)=\mathrm{0} \\$$$$\frac{\mathrm{df}}{\mathrm{dy}}=\mathrm{2x}−\mathrm{8}+\lambda\left(\mathrm{2y}\right)=\mathrm{0} \\$$$$\frac{\mathrm{df}}{\mathrm{d}\lambda}=\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{50}=\mathrm{0} \\$$

Commented by mind is power last updated on 23/Jan/20

$${nice}\:{But}\:{your}\:{approch}\:{didn}'{t}\:{use}\:{the}\:{contraint}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{50} \\$$$${try}\:{to}\:{put}\:\:{g}\left({x},{y},\gamma\right)=\mathrm{2}{xy}−\mathrm{8}{x}−\mathrm{8}{y}+\mathrm{70}+\gamma\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{50}\right) \\$$$$\\$$

Commented by jagoll last updated on 23/Jan/20

$$\mathrm{oo}\:\mathrm{yes}\:.\:\mathrm{it}\:\mathrm{Langrange}\:\mathrm{method} \\$$

Commented by jagoll last updated on 23/Jan/20

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\$$

Commented by jagoll last updated on 23/Jan/20

$$\mathrm{that}\:\mathrm{right}\:\mathrm{sir}? \\$$

Commented by jagoll last updated on 23/Jan/20

$$\frac{\mathrm{2}\lambda\mathrm{x}}{\mathrm{2}\lambda\mathrm{y}}=\frac{\mathrm{8}−\mathrm{2y}}{\mathrm{8}−\mathrm{2x}\:}\Rightarrow\frac{\mathrm{x}}{\mathrm{y}}=\frac{\mathrm{4}−\mathrm{y}}{\mathrm{4}−\mathrm{x}} \\$$$$\mathrm{4x}−\mathrm{x}^{\mathrm{2}} =\mathrm{4y}−\mathrm{y}^{\mathrm{2}} \\$$$$\left(\mathrm{x}−\mathrm{2}\right)^{\mathrm{2}} =\left(\mathrm{y}−\mathrm{2}\right)^{\mathrm{2}} \\$$$$\mathrm{x}−\mathrm{2}=\pm\:\left(\mathrm{y}−\mathrm{2}\right)\:\Rightarrow\mathrm{x}=\mathrm{2}\pm\left(\mathrm{y}−\mathrm{2}\right) \\$$$$\mathrm{x}_{\mathrm{1}} \:=\mathrm{y}\:\vee\:\mathrm{x}_{\mathrm{2}} =\mathrm{4}−\mathrm{y} \\$$

Commented by jagoll last updated on 23/Jan/20

$$\mathrm{case}\:\left(\mathrm{1}\right)\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{50}\Rightarrow\mathrm{x}=\pm\mathrm{5}=\mathrm{y} \\$$$$\mathrm{f}=\left(\mathrm{10}\right)^{\mathrm{2}} −\mathrm{8}\left(\mathrm{10}\right)+\mathrm{20}=\mathrm{40} \\$$

Commented by jagoll last updated on 23/Jan/20

$$\mathrm{case}\left(\mathrm{2}\right)\:\left(\mathrm{4}−\mathrm{y}\right)^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{50}=\mathrm{0} \\$$$$\mathrm{2y}^{\mathrm{2}} −\mathrm{8y}−\mathrm{34}=\mathrm{0} \\$$$$\mathrm{y}^{\mathrm{2}} −\mathrm{4y}−\mathrm{17}=\mathrm{0} \\$$$$\left(\mathrm{y}−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{21}=\mathrm{0}\:\Rightarrow\mathrm{y}=\mathrm{2}+\sqrt{\mathrm{21}} \\$$

Commented by jagoll last updated on 23/Jan/20

$$\mathrm{mr}\:\mathrm{W}\:\mathrm{what}\:\mathrm{the}\:\mathrm{answer}? \\$$

Commented by mind is power last updated on 23/Jan/20

$${Nice}\:{Sir} \\$$

Answered by mind is power last updated on 23/Jan/20

$${x}=\sqrt{\mathrm{50}}{cos}\left(\theta\right) \\$$$${y}=\sqrt{\mathrm{50}}{sin}\left(\theta\right) \\$$$${x}+{y}=\sqrt{\mathrm{50}}.\left({cos}\left(\theta\right)+{sin}\left(\theta\right)\right) \\$$$$=\mathrm{10}{sin}\left(\theta+\frac{\pi}{\mathrm{4}}\right)=\mathrm{10}{sin}\left(\varphi\right),\varphi\in\left[\mathrm{0},\mathrm{2}\pi\left[\right.\right. \\$$$${x}+{y}\in\left[−\mathrm{10},\mathrm{10}\right] \\$$$$\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{8}\left({x}+{y}\right)+\mathrm{20}={f}\left({x}+{y}\right) \\$$$${f}\left({t}\right)={t}^{\mathrm{2}} −\mathrm{8}{t}+\mathrm{20} \\$$$${t}\in\left[−\mathrm{10},\mathrm{10}\right] \\$$$${f}'\left({t}\right)=\mathrm{2}{t}−\mathrm{8},{minf}={f}\left(\mathrm{4}\right)=\mathrm{16}−\mathrm{32}+\mathrm{20}=\mathrm{4} \\$$$${x}+{y}=\mathrm{4}=\mathrm{10}{sin}\left(\varphi\right),\varphi=\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{5}}\right),\theta=\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{5}}\right)−\frac{\pi}{\mathrm{4}} \\$$$${x}=\sqrt{\mathrm{50}}{cos}\left(\theta\right),{y}=\sqrt{\mathrm{50}}{sin}\left(\theta\right) \\$$$${mox}\:{f}\left(−\mathrm{10}\right)=\mathrm{100}+\mathrm{80}+\mathrm{20}=\mathrm{200} \\$$$$−\mathrm{10}=\mathrm{10}{sin}\left(\varphi\right)\Rightarrow\varphi=−\frac{\pi}{\mathrm{2}}\Rightarrow\theta=−\frac{\mathrm{3}\pi}{\mathrm{4}} \\$$$${x}=\sqrt{\mathrm{50}}.−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}=−\mathrm{5}\:\:,{y}=−\mathrm{5} \\$$$$\\$$

Answered by mr W last updated on 23/Jan/20

$${it}\:{is}\:{to}\:{find}\:{the}\:{minimum}\:{and}\: \\$$$${maximum}\:{of}\:{f}=\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{8}\left({x}+{y}\right)+\mathrm{20} \\$$$${under}\:{the}\:{condition}\:{that}\:{x},\:{y}\:{satisfy} \\$$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{50},\:{i}.{e}.\:\left({x},\:{y}\right)\:{is}\:{on}\:{the}\:{circle}. \\$$$${let}\:{x}=\mathrm{5}\sqrt{\mathrm{2}}\:\mathrm{cos}\:\theta,\:{y}=\mathrm{5}\sqrt{\mathrm{2}}\:\mathrm{sin}\:\theta \\$$$${f}=\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{8}\left({x}+{y}\right)+\mathrm{20} \\$$$$=\left({x}+{y}−\mathrm{4}\right)^{\mathrm{2}} +\mathrm{4} \\$$$$=\left(\mathrm{5}\sqrt{\mathrm{2}}\:\mathrm{cos}\:\theta+\mathrm{5}\sqrt{\mathrm{2}}\:\mathrm{sin}\:\theta−\mathrm{4}\right)^{\mathrm{2}} +\mathrm{4} \\$$$$=\left[\mathrm{10}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:\mathrm{cos}\:\theta+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:\mathrm{sin}\:\theta\right)−\mathrm{4}\right]^{\mathrm{2}} +\mathrm{4} \\$$$$=\left[\mathrm{10}\:\mathrm{cos}\:\left(\theta−\frac{\pi}{\mathrm{4}}\right)−\mathrm{4}\right]^{\mathrm{2}} +\mathrm{4}\geqslant\mathrm{4} \\$$$${f}_{{min}} \:{is}\:{when}\:\mathrm{cos}\:\left(\theta−\frac{\pi}{\mathrm{4}}\right)=\frac{\mathrm{4}}{\mathrm{10}},\:{i}.{e}.\:\theta=\frac{\pi}{\mathrm{4}}+\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{2}}{\mathrm{5}}, \\$$$${f}_{{min}} =\left(\mathrm{4}−\mathrm{4}\right)^{\mathrm{2}} +\mathrm{4}=\mathrm{4} \\$$$${f}_{{max}} \:{is}\:{when}\:\mathrm{cos}\:\left(\theta−\frac{\pi}{\mathrm{4}}\right)=−\mathrm{1},\:{i}.{e}.\:\theta=\frac{\mathrm{5}\pi}{\mathrm{4}}, \\$$$${f}_{{max}} =\left(−\mathrm{10}−\mathrm{4}\right)^{\mathrm{2}} +\mathrm{4}=\mathrm{200} \\$$

Answered by key of knowledge last updated on 23/Jan/20

$$\mathrm{u}=\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{8}\left({x}+{y}\right)+\mathrm{20}=\left(\left(\mathrm{x}+\mathrm{y}\right)−\mathrm{4}\right)^{\mathrm{2}} +\mathrm{4} \\$$$$\mathrm{if}\left(\left(\mathrm{x}+\mathrm{y}\right)−\mathrm{4}\right)=\mathrm{0}\Rightarrow\mathrm{min}\left(\mathrm{u}\right)=\mathrm{4} \\$$$$\mathrm{min}\left(\mathrm{x}+\mathrm{y}\right)=−\mathrm{10}\:\&\:\mathrm{max}\left(\mathrm{x}+\mathrm{y}\right)=\mathrm{10} \\$$$$\mathrm{ax}\left(\left(\mathrm{x}+\mathrm{y}\right)−\mathrm{4}\right)=\left(−\mathrm{14}\right)\Rightarrow\mathrm{max}\left(\mathrm{u}\right)=\mathrm{200} \\$$

Answered by john santu last updated on 24/Jan/20

$${because}\:{x}\:{and}\:{y}\:{meet}\:{the}\: \\$$$${constraints}\:{of}\:{circle}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{50} \\$$$$,\:{let}\:{x}+{y}\:=\:{k}\:{is}\:{tangent}\:{line}\:\:{circle}\:, \\$$$${then}\:{distance}\:{of}\:{the}\:{center}\: \\$$$${of}\:{the}\:{circle}\:{to}\:{tangent}\:{is}\:{the} \\$$$${radius}\:\Rightarrow{r}=\mid\frac{{k}}{\sqrt{\mathrm{2}}}\mid\:,\mid{k}\mid=\:\mathrm{10}\:\Rightarrow{k}\:=\pm\mathrm{10} \\$$$${for}\:{x}+{y}\:=\:−\mathrm{10}\:\Rightarrow{f}=\left(−\mathrm{10}\right)^{\mathrm{2}} −\mathrm{8}\left(−\mathrm{10}\right)+\mathrm{20}=\mathrm{200} \\$$$$\therefore{f}_{{max}} =\mathrm{200}\: \\$$$${consider}\:{x}+{y}\:={t} \\$$$${f}\left({t}\right)\:={t}^{\mathrm{2}} −\mathrm{8}{t}+\mathrm{20}\:,\:{a}\:{quadratic} \\$$$${function}\:{in}\:{t}.\:{the}\:{minimum} \\$$$${value}\:{when}\:{t}\:=\:\frac{−{b}}{\mathrm{2}{a}}=\frac{−\left(−\mathrm{8}\right)}{\mathrm{2}.\mathrm{1}}=\mathrm{4} \\$$$${the}\:{f}_{{min}} \:=\:\mathrm{4}^{\mathrm{2}} −\mathrm{8}×\mathrm{4}+\mathrm{20}=\mathrm{4} \\$$$$\\$$

Commented by mr W last updated on 23/Jan/20

$${f}_{{min}} \neq\mathrm{40}\:{but}\:=\mathrm{4}. \\$$