plz solve this prob and tell me

No answers to this one??

let the sides be a,a,b

then

2a+b=P=2s where s is the semiperimeter --- (i)

differentiating both sides w.r.t b

2da/db + 1=2ds/db=dP/db ------- (ii)

area of the triangle=rs where r given is the inradius
also area=√(s(s-a)(s-a)(s-b))

so

r²s²=s(s-a)²(s-b) ------(iii)

hence as we know a as a function of b s will come as a function of b too

at maximum perimeter

dP/db=0

hence d(2s)/db=0

ds/db=0 ---------(v)

thus from (ii)

da/db=-1/2

further from (iii)

differentiating it w.r.t b
d/db(r²s)=d/db((s-a)²(s-b))

at maximum perimeter ds/db=0

hence we get

2(s-a)(s-b)(ds/db-da/db)+ (s-a)²(ds/db-1)=0

from(v)

da/db=-1/2

hence

2(s-a)(s-b)(1/2) + (s-a)²(-1)=0

we get

s-a=s-b

or a=b

thus maximum perimeter in case of an equilateral triangle ..

side of triangle=

2(r/tan30)=2√3r

perimeter= side*3=6√3r

any conformations to the above .. or any shorter method?