Question Number 74713 by ajfour last updated on 29/Nov/19
Commented by ajfour last updated on 29/Nov/19
$${If}\:{perimeter}\:{of}\:\bigtriangleup{ABC}\:=\:\mathrm{18}\:, \\ $$$${and}\:{area}\:{is}\:{maximum},\:{find} \\ $$$${coordinates}\:{of}\:{points}\:{A},\:{B},\:{C}. \\ $$
Answered by mr W last updated on 29/Nov/19
$${AB}={BC}={CA}=\mathrm{6} \\ $$$${y}_{{C}} =\mathrm{3}\sqrt{\mathrm{3}} \\ $$$${x}_{{C}} =\frac{\mathrm{9}\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$$${x}_{{A}} =\frac{\mathrm{9}\sqrt{\mathrm{3}}}{\mathrm{4}}−\mathrm{3} \\ $$$${x}_{{B}} =\frac{\mathrm{9}\sqrt{\mathrm{3}}}{\mathrm{4}}+\mathrm{3} \\ $$
Commented by mr W last updated on 29/Nov/19
$${this}\:{is}\:{only}\:{because}\:\alpha\leqslant\mathrm{60}° \\ $$$${for}\:\alpha>\mathrm{60}°\:{it}'{s}\:{a}\:{different}\:{triangle}\:{shape}. \\ $$
Commented by ajfour last updated on 29/Nov/19
$${Its}\:{alright}\:{Sir},\:{I}\:{dint}\:{think}\:{much}. \\ $$
Commented by mr W last updated on 29/Nov/19
Commented by mr W last updated on 30/Nov/19
$${for}\:\alpha>\mathrm{60}° \\ $$$${case}\:\mathrm{1}: \\ $$$${AC}={BC}=\frac{{AB}}{\mathrm{2}\:\mathrm{cos}\:\alpha} \\ $$$${AB}+\frac{{AB}}{\mathrm{cos}\:\alpha}={p}=\mathrm{18} \\ $$$${AB}=\frac{{p}\:\mathrm{cos}\:\alpha}{\mathrm{1}+\mathrm{cos}\:\alpha} \\ $$$${Area}_{\mathrm{1}} =\frac{{AB}^{\mathrm{2}} \mathrm{tan}\:\alpha}{\mathrm{2}}=\frac{{p}^{\mathrm{2}} \mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha}{\mathrm{2}\left(\mathrm{1}+\mathrm{cos}\:\alpha\right)^{\mathrm{2}} } \\ $$$${case}\:\mathrm{2}: \\ $$$${AB}={AC} \\ $$$${BC}={AB}\:\mathrm{sin}\:\frac{\alpha}{\mathrm{2}} \\ $$$$\mathrm{2}{AB}+{AB}\:\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}={p}=\mathrm{18} \\ $$$${AB}=\frac{{p}}{\mathrm{2}+\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}} \\ $$$${Area}_{\mathrm{2}} =\frac{{AB}^{\mathrm{2}} \mathrm{sin}\:\alpha}{\mathrm{2}}=\frac{{p}^{\mathrm{2}} \mathrm{sin}\:\alpha}{\mathrm{2}\left(\mathrm{2}+\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$\frac{{Area}_{\mathrm{1}} }{{Area}_{\mathrm{2}} }=\mathrm{cos}\:\alpha\left(\frac{\mathrm{2}+\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}}{\mathrm{1}+\mathrm{cos}\:\alpha}\right)^{\mathrm{2}} \\ $$$${if}\:\alpha\leqslant\mathrm{77}.\mathrm{68}°\:{case}\:\mathrm{1}\:{gives}\:{maximum}\:{area}. \\ $$$${if}\:\alpha>\mathrm{77}.\mathrm{68}°\:{case}\:\mathrm{2}\:{gives}\:{maximum}\:{area}. \\ $$