Question Number 39868 by math2018 last updated on 12/Jul/18 | ||
$${a},{b},{c}>\mathrm{0},{if} \\ $$ $$\frac{\mathrm{8}{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} +\mathrm{9}}={b},\:\:\frac{\mathrm{10}{b}^{\mathrm{2}} }{{b}^{\mathrm{2}} +\mathrm{16}}={c},\:\:\frac{\mathrm{6}{c}^{\mathrm{2}} }{{c}^{\mathrm{2}} +\mathrm{25}}={a}, \\ $$ $${then},{what}\:{is}\:{the}\:{minimum}\:{value}\:{of}\:\:{a}+{b}+{c}? \\ $$ | ||
Answered by tanmay.chaudhury50@gmail.com last updated on 12/Jul/18 | ||
$${b}=\frac{\mathrm{8}}{\mathrm{1}+\frac{\mathrm{9}}{{a}^{\mathrm{2}} }}=\frac{\mathrm{8}}{\left(\mathrm{1}−\frac{\mathrm{3}}{{a}}\right)^{\mathrm{2}} +\frac{\mathrm{6}}{{a}}} \\ $$ $${if}\:\:{a}=\mathrm{3}\:\:{then}\:\:\:{b}=\mathrm{4} \\ $$ $${c}=\frac{\mathrm{10}}{\left(\mathrm{1}−\frac{\mathrm{4}}{{b}}\right)^{\mathrm{2}} +\frac{\mathrm{8}}{{b}}}\:\:{if}\:{b}=\mathrm{4}\:\:\:\:{c}=\mathrm{5} \\ $$ $${a}=\frac{\mathrm{6}}{\left(\mathrm{1}−\frac{\mathrm{5}}{{c}}\right)^{\mathrm{2}} +\frac{\mathrm{10}}{{c}}}\:\:{if}\:{c}=\mathrm{5}\:\:\:{a}=\mathrm{3} \\ $$ $${so}\:{a}=\mathrm{3} \\ $$ $${b}=\mathrm{4} \\ $$ $${c}=\mathrm{5}\:{satisfy}\:{three}\:{eqn} \\ $$ $${a}+{b}+{c}=\mathrm{12} \\ $$ $${pls}\:{check}...{whether}\:{any}\:{direct}\:{method}\:{to}\:{solve} \\ $$ $${i}\:{do}\:{not}\:{know}... \\ $$ $$ \\ $$ | ||
Commented bymath2018 last updated on 13/Jul/18 | ||
$${Thank}\:{you}\:{for}\:{your}\:{work}! \\ $$ | ||
Answered by ajfour last updated on 12/Jul/18 | ||
$${let}\:\boldsymbol{{a}}=\mathrm{3tan}\:\boldsymbol{\alpha},\:\boldsymbol{{b}}=\mathrm{4tan}\:\boldsymbol{\beta},\:\boldsymbol{{c}}=\mathrm{5tan}\:\boldsymbol{\gamma} \\ $$ $${Then} \\ $$ $$\frac{\mathrm{8}×\mathrm{9tan}\:^{\mathrm{2}} \alpha}{\mathrm{9sec}\:^{\mathrm{2}} \alpha}\:=\mathrm{4tan}\:\beta \\ $$ $$\Rightarrow\:\:\mathrm{2sin}\:^{\mathrm{2}} \alpha\:=\:\mathrm{tan}\:\beta\:\:\:\:\:\:\:....\left({i}\right) \\ $$ $$\frac{\mathrm{10}×\mathrm{16tan}\:^{\mathrm{2}} \beta}{\mathrm{16sec}\:^{\mathrm{2}} \beta}\:=\mathrm{5tan}\:\gamma \\ $$ $$\Rightarrow\:\:\mathrm{2sin}\:^{\mathrm{2}} \beta\:=\:\mathrm{tan}\:\gamma\:\:\:\:\:\:....\left({ii}\right) \\ $$ $$\frac{\mathrm{6}×\mathrm{25tan}\:^{\mathrm{2}} \gamma}{\mathrm{25sec}\:^{\mathrm{2}} \gamma}\:=\:\mathrm{3tan}\:\alpha \\ $$ $$\Rightarrow\:\:\mathrm{2sin}\:^{\mathrm{2}} \gamma\:=\:\mathrm{tan}\:\alpha\:\:\:\:\:\:\:\:....\left({iii}\right) \\ $$ $$\left({i}\right)×\left({ii}\right)×\left({iii}\right)\:\:\:{gives} \\ $$ $$\mathrm{8sin}\:^{\mathrm{2}} \alpha\:\mathrm{sin}\:^{\mathrm{2}} \beta\:\mathrm{sin}\:^{\mathrm{2}} \gamma\:=\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta\:\mathrm{tan}\:\gamma \\ $$ $${or}\:\:\:\:\mathrm{sin}\:\mathrm{2}\alpha\:\mathrm{sin}\:\mathrm{2}\beta\:\mathrm{sin}\:\mathrm{2}\gamma\:=\:\mathrm{1} \\ $$ $$\Rightarrow\:\:\alpha\:=\:\beta\:=\:\gamma\:=\:\frac{\pi}{\mathrm{4}} \\ $$ $${a}+{b}+{c}\:=\:\mathrm{3tan}\:\alpha+\mathrm{4tan}\:\beta+\mathrm{5tan}\:\gamma \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left(\mathrm{3}+\mathrm{4}+\mathrm{5}\right)\mathrm{tan}\:\frac{\pi}{\mathrm{4}} \\ $$ $${a}+{b}+{c}\:=\:\mathrm{12}\:. \\ $$ $$\left({why}\:{the}\:{minimum}?\:\right)! \\ $$ | ||
Commented bytanmay.chaudhury50@gmail.com last updated on 13/Jul/18 | ||
$${yes}\:\:{i}\:{forgot}\:{it}\:...{thank}\:{you}\:{for}\:{correction} \\ $$ | ||
Commented bymath2018 last updated on 13/Jul/18 | ||
$$\because{a}>\mathrm{0},{b}>\mathrm{0},{c}>\mathrm{0} \\ $$ $$\therefore{a}+{b}+{c}\neq\mathrm{0} \\ $$ $${Thank}\:{you}\:{Sir}. \\ $$ | ||
Commented bymath2018 last updated on 13/Jul/18 | ||
$${a}=\frac{\mathrm{6}{c}^{\mathrm{2}} }{{c}^{\mathrm{2}} +\mathrm{25}}\leqslant\frac{\mathrm{6}{c}^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{25}{c}^{\mathrm{2}} }}=\frac{\mathrm{3}{c}}{\mathrm{5}} \\ $$ $${b}=\frac{\mathrm{8}{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} +\mathrm{9}}\leqslant\frac{\mathrm{8}{a}^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{9}{a}^{\mathrm{2}} }}=\frac{\mathrm{4}{a}}{\mathrm{3}} \\ $$ $${c}=\frac{\mathrm{10}{b}^{\mathrm{2}} }{{b}^{\mathrm{2}} +\mathrm{16}}\leqslant\frac{\mathrm{10}{b}^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{16}{b}^{\mathrm{2}} }}=\frac{\mathrm{5}{b}}{\mathrm{4}} \\ $$ $${now}\:{we}\:{can}\:{get}\:{the}\:{minimum}\:{value}\:{of}\:{a}+{b}+{c} \\ $$ $${Thank}\:{you}\:{very}\:{much}! \\ $$ | ||
Commented bytanmay.chaudhury50@gmail.com last updated on 13/Jul/18 | ||
$$\frac{{a}}{\mathrm{3}}=\frac{{b}}{\mathrm{4}}=\frac{{c}}{\mathrm{5}}={k} \\ $$ $${a}=\mathrm{3}{k}\:\:{b}=\mathrm{4}{k}\:\:{c}=\mathrm{5}{k}\:\: \\ $$ $${a}=\frac{\mathrm{6}{c}^{\mathrm{2}} }{{c}^{\mathrm{2}} +\mathrm{25}}=\frac{\mathrm{6}×\mathrm{25}{k}^{\mathrm{2}} }{\mathrm{25}{k}^{\mathrm{2}} +\mathrm{25}} \\ $$ $$\mathrm{3}{k}=\frac{\mathrm{6}{k}^{\mathrm{2}} }{{k}^{\mathrm{2}} +\mathrm{1}} \\ $$ $$\mathrm{3}{k}\left({k}^{\mathrm{2}} +\mathrm{1}\right)−\mathrm{6}{k}^{\mathrm{2}} =\mathrm{0} \\ $$ $$\mathrm{3}{k}\left({k}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}{k}\right)=\mathrm{0} \\ $$ $${so}\:{k}=\mathrm{0}\:{and}\:{k}=\mathrm{1} \\ $$ $${a}+{b}+{c} \\ $$ $$=\mathrm{3}{k}+\mathrm{4}{k}+\mathrm{5}{k}=\mathrm{12}{k} \\ $$ $${so}\:{when}\:{k}=\mathrm{0}\:\:\:{a}+{b}+{c}=\mathrm{0} \\ $$ $${when}\:{k}=\mathrm{1}\:\:{a}+{b}+{c}=\mathrm{12} \\ $$ $${hence}\:{min}\:{value}\:{is}\:{zero}...{pls}\:{check}... \\ $$ $$ \\ $$ | ||