Calculus 1
posted by Oswaldo .
find the area of the largest posible isosceles triangle with 2 sides equal to 6. thanks

hhight , athirth side of triangle
alangle ALPHA , Aarea
sin(al)=(h/6) , h=6*sin(al)
cos(al)=(a/2)/6 =( a/12) , a=12*cos(al)
A=(1/2)a*h=(1/2)*6*sin(al)*12*cos(al)
=(1/2)*72*sin(al)*cos(al) =36*sin(al)*cos(al) =18*(2*sin(al)*cos(al))=18*sin(2al)
=dA/dal)=18*2*cos(2al=0 , cos(2al)=0
2al=90° al=45° 
h=6*sin(45‹)=6*(1/ã2)=6/ã2
a=12*12*cos(45)=12*(1/ã2)=12/ã2
A=(1/2)a*h=(1/2)6/ã2*12/ã2
=(1/2)*(1/ã2)/*1/ã2)*72=(1/2)*(1/)*72
=(1/4)*72= 18
Largest posible area= 18 
ã2 is square root
Area=(1/2)*(1/2)*72=18 
thanks