Question-74713

Question Number 74713 by ajfour last updated on 29/Nov/19

Commented by ajfour last updated on 29/Nov/19

If perimeter of △ABC = 18 , and area is maximum, find coordinates of points A, B, C.

$${If}\:{perimeter}\:{of}\:\bigtriangleup{ABC}\:=\:\mathrm{18}\:, \\ $$$${and}\:{area}\:{is}\:{maximum},\:{find} \\ $$$${coordinates}\:{of}\:{points}\:{A},\:{B},\:{C}. \\ $$

Answered by mr W last updated on 29/Nov/19

AB=BC=CA=6 y_C =3(√3) x_C =((9(√3))/4) x_A =((9(√3))/4)−3 x_B =((9(√3))/4)+3

$${AB}={BC}={CA}=\mathrm{6} \\ $$$${y}_{{C}} =\mathrm{3}\sqrt{\mathrm{3}} \\ $$$${x}_{{C}} =\frac{\mathrm{9}\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$$${x}_{{A}} =\frac{\mathrm{9}\sqrt{\mathrm{3}}}{\mathrm{4}}−\mathrm{3} \\ $$$${x}_{{B}} =\frac{\mathrm{9}\sqrt{\mathrm{3}}}{\mathrm{4}}+\mathrm{3} \\ $$

Commented by mr W last updated on 29/Nov/19

this is only because α≤60° for α>60° it′s a different triangle shape.

$${this}\:{is}\:{only}\:{because}\:\alpha\leqslant\mathrm{60}° \\ $$$${for}\:\alpha>\mathrm{60}°\:{it}'{s}\:{a}\:{different}\:{triangle}\:{shape}. \\ $$

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 α>60° case 1: AC=BC=((AB)/(2 cos α)) AB+((AB)/(cos α))=p=18 AB=((p cos α)/(1+cos α)) Area_1 =((AB^2 tan α)/2)=((p^2 sin α cos α)/(2(1+cos α)^2 )) case 2: AB=AC BC=AB sin (α/2) 2AB+AB sin (α/2)=p=18 AB=(p/(2+sin (α/2))) Area_2 =((AB^2 sin α)/2)=((p^2 sin α)/(2(2+sin (α/2))^2 )) ((Area_1 )/(Area_2 ))=cos α(((2+sin (α/2))/(1+cos α)))^2 if α≤77.68° case 1 gives maximum area. if α>77.68° case 2 gives maximum area.

$${for}\:\alpha>\mathrm{60}° \\ $$$${case}\:\mathrm{1}: \\ $$$${AC}={BC}=\frac{{AB}}{\mathrm{2}\:\mathrm{cos}\:\alpha} \\ $$$${AB}+\frac{{AB}}{\mathrm{cos}\:\alpha}={p}=\mathrm{18} \\ $$$${AB}=\frac{{p}\:\mathrm{cos}\:\alpha}{\mathrm{1}+\mathrm{cos}\:\alpha} \\ $$$${Area}_{\mathrm{1}} =\frac{{AB}^{\mathrm{2}} \mathrm{tan}\:\alpha}{\mathrm{2}}=\frac{{p}^{\mathrm{2}} \mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha}{\mathrm{2}\left(\mathrm{1}+\mathrm{cos}\:\alpha\right)^{\mathrm{2}} } \\ $$$${case}\:\mathrm{2}: \\ $$$${AB}={AC} \\ $$$${BC}={AB}\:\mathrm{sin}\:\frac{\alpha}{\mathrm{2}} \\ $$$$\mathrm{2}{AB}+{AB}\:\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}={p}=\mathrm{18} \\ $$$${AB}=\frac{{p}}{\mathrm{2}+\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}} \\ $$$${Area}_{\mathrm{2}} =\frac{{AB}^{\mathrm{2}} \mathrm{sin}\:\alpha}{\mathrm{2}}=\frac{{p}^{\mathrm{2}} \mathrm{sin}\:\alpha}{\mathrm{2}\left(\mathrm{2}+\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$\frac{{Area}_{\mathrm{1}} }{{Area}_{\mathrm{2}} }=\mathrm{cos}\:\alpha\left(\frac{\mathrm{2}+\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}}{\mathrm{1}+\mathrm{cos}\:\alpha}\right)^{\mathrm{2}} \\ $$$${if}\:\alpha\leqslant\mathrm{77}.\mathrm{68}°\:{case}\:\mathrm{1}\:{gives}\:{maximum}\:{area}. \\ $$$${if}\:\alpha>\mathrm{77}.\mathrm{68}°\:{case}\:\mathrm{2}\:{gives}\:{maximum}\:{area}. \\ $$

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