Question Number 38391 by gunawan last updated on 25/Jun/18
$${a}+\mathrm{2}{b}+\mathrm{3}{c}=\mathrm{12} \\ $$$$\mathrm{2}{ab}+\mathrm{3}{ac}+\mathrm{6}{bc}=\mathrm{48} \\ $$$${a}+{b}+{c}=... \\ $$
Answered by MrW3 last updated on 26/Jun/18
$${let} \\ $$$${x}={a} \\ $$$${y}=\mathrm{2}{b} \\ $$$${z}=\mathrm{3}{c} \\ $$$$ \\ $$$$\left({i}\right)\:\Rightarrow{x}+{y}+{z}=\mathrm{12}\Rightarrow{z}=\mathrm{12}−{x}−{y} \\ $$$$\left({ii}\right)\:\Rightarrow{xy}+{yz}+{zx}=\mathrm{48}\Rightarrow{xy}+\mathrm{12}\left({x}+{y}\right)−\left({x}+{y}\right)^{\mathrm{2}} =\mathrm{48} \\ $$$$ \\ $$$${F}\left({x},{y}\right)={xy}+\mathrm{12}\left({x}+{y}\right)−\left({x}+{y}\right)^{\mathrm{2}} \\ $$$$\frac{\partial{F}}{\partial{x}}={y}+\mathrm{12}−\mathrm{2}\left({x}+{y}\right)=\mathrm{0}\Rightarrow\mathrm{2}{x}+{y}=\mathrm{12} \\ $$$$\frac{\partial{F}}{\partial{y}}={x}+\mathrm{12}−\mathrm{2}\left({x}+{y}\right)=\mathrm{0}\Rightarrow{x}+\mathrm{2}{y}=\mathrm{12} \\ $$$$\begin{cases}{\mathrm{2}{x}+{y}=\mathrm{12}}\\{{x}+\mathrm{2}{y}=\mathrm{12}}\end{cases}\Rightarrow{x}={y}=\mathrm{4}\Rightarrow{F}\left(\mathrm{4},\mathrm{4}\right)=\mathrm{48} \\ $$$$\Rightarrow{maximum}\:{from}\:{F}\left({x},{y}\right)\:{is}\:{at}\:{x}={y}=\mathrm{4} \\ $$$${which}\:{is}\:\mathrm{48},\:{that}\:{is}\:{to}\:{say} \\ $$$$\begin{cases}{{x}+{y}+{z}=\mathrm{12}}\\{{xy}+{yz}+{zx}=\mathrm{48}}\end{cases} \\ $$$${has}\:{one}\:{and}\:{only}\:{one}\:{solution}: \\ $$$${x}={y}={z}=\mathrm{4} \\ $$$$\Rightarrow{a}=\mathrm{4},\:{b}=\frac{\mathrm{4}}{\mathrm{2}}=\mathrm{2},\:{c}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\Rightarrow{a}+{b}+{c}=\mathrm{4}+\mathrm{2}+\frac{\mathrm{4}}{\mathrm{3}}=\frac{\mathrm{22}}{\mathrm{3}} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 25/Jun/18
$${excellent}... \\ $$
Commented by gunawan last updated on 26/Jun/18
$$\mathrm{wow}\:\mathrm{nice}\:\mathrm{Sir} \\ $$