Question Number 143234 by mathdanisur last updated on 11/Jun/21
Answered by mr W last updated on 11/Jun/21
$${F}={x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} +\lambda\left(\alpha{x}+\alpha{y}+\gamma{z}−\mathrm{1}\right) \\ $$$$\frac{\partial{F}}{\partial{x}}=\mathrm{3}{x}^{\mathrm{2}} +\lambda\alpha=\mathrm{0}\: \\ $$$$\Rightarrow{x}=\sqrt{−\frac{\lambda\alpha}{\mathrm{3}}} \\ $$$${similarly} \\ $$$$\Rightarrow{y}=\sqrt{−\frac{\lambda\beta}{\mathrm{3}}} \\ $$$$\Rightarrow{z}=\sqrt{−\frac{\lambda\gamma}{\mathrm{3}}} \\ $$$$\Rightarrow\left(\alpha\sqrt{\alpha}+\beta\sqrt{\beta}+\gamma\sqrt{\gamma}\right)\sqrt{−\frac{\lambda}{\mathrm{3}}}=\mathrm{1} \\ $$$$\Rightarrow\sqrt{−\frac{\lambda}{\mathrm{3}}}=\frac{\mathrm{1}}{\alpha\sqrt{\alpha}+\beta\sqrt{\beta}+\gamma\sqrt{\gamma}} \\ $$$$\Rightarrow{x}=\frac{\sqrt{\alpha}}{\alpha\sqrt{\alpha}+\beta\sqrt{\beta}+\gamma\sqrt{\gamma}} \\ $$$$\Rightarrow{y}=\frac{\sqrt{\beta}}{\alpha\sqrt{\alpha}+\beta\sqrt{\beta}+\gamma\sqrt{\gamma}} \\ $$$$\Rightarrow{z}=\frac{\sqrt{\gamma}}{\alpha\sqrt{\alpha}+\beta\sqrt{\beta}+\gamma\sqrt{\gamma}} \\ $$$${P}_{{min}} =\frac{\alpha\sqrt{\alpha}+\beta\sqrt{\beta}+\gamma\sqrt{\gamma}}{\left(\alpha\sqrt{\alpha}+\beta\sqrt{\beta}+\gamma\sqrt{\gamma}\right)^{\mathrm{3}} }=\frac{\mathrm{1}}{\left(\alpha\sqrt{\alpha}+\beta\sqrt{\beta}+\gamma\sqrt{\gamma}\right)^{\mathrm{2}} } \\ $$
Commented by mathdanisur last updated on 12/Jun/21
$${thaks}\:{Sir},\:{sorry}\:{please}\:{check}\:{again}.. \\ $$
Commented by mr W last updated on 12/Jun/21
$${what}\:{shall}\:{i}\:{check}?\:{is}\:{my}\:{answer} \\ $$$${wrong}? \\ $$
Commented by mathdanisur last updated on 12/Jun/21
$${yes}\:{Sir},\:{please}\:{solve}\:{again}.. \\ $$
Commented by mr W last updated on 12/Jun/21
$${i}\:{can}'{t}\:{find}\:{mistake}\:{in}\:{my}\:{answer}. \\ $$$${if}\:{it}'{s}\:{wrong},\:{please}\:{tell}\:{me}\:{the}\:{right} \\ $$$${answer}. \\ $$
Commented by mathdanisur last updated on 13/Jun/21
$${thank}\:{you}\:{so}\:{much}\:{dear}\:{Sir}.. \\ $$