How to calculate the maximum area of an isosceles triangle given only perimeter?

ABC isosceles by A

BC=x

AB+AC+BC = 27
AB+AC + x = 27
AB+AC = 27 - x
but AB=AC so
2AB = 27 - x
AB = (27 - x) / 2 = AC

(AH) _|_ (BC) (perpendicular)
H (is in) [BC]

That is on say (AH) is the height of [BC] that develop from A
As ABC is isosceles, (AH) is and the median of [BC]

So BH= BC / 2 = x/2

AHB is right-angles by H. Pythagore says :

AB^2 = AH^2 + BH^2

[ (27-x)/2 ]^2= AH^2 + (x^2)/ 4

AH^2 = [ (27-x)/2 ]^2 - (x^2)/ 4

AH^2 = [729 - 54x+x^2 - x^2 ] / 4

AH^2 = (729 - 54x) / 4

AH = V(729 - 54x) / 2


Area= V(729 - 54x) * x / 4

Area=f(x)

f(x)=V(729 - 54x) * x / 4

f(x)= (1 / 4) * V(729 - 54x) * x

differentiate :

f '(x) = (1 / 4) * [ - 54* 1 / [ 2* V(729 - 54x) ] * x + V(729 - 54x) * 1 ]


formula : (uv)' = u'v + uv'

f '(x) = (1 / 4) * [ -54x / [ 2* V(729 - 54x) ] + V(729 - 54x) ]
f '(x) = (1 / 4) * [ -27x / V(729 - 54x) + V(729 - 54x)^2/ V(729 - 54x)]
f '(x) = (1 / 4) * [ (-27x + 729 -54x) / V(729 - 54x) ]
f' (x)= (-81x + 729)/ [ 4 * V(729 - 54x) ]

NB:
V(729 - 54x) =/= 0 so (729 - 54x) =/= 0
(729 - 54x) > 0
x < 13.5

And x >= 0

so x (in in) [ 0 ; 13.5 [

[ 4 * V(729 - 54x) ] > 0 so the sign connected with f'(x) is the sign of

(-81x + 729).

-81x + 729 > 0
when x < 9

so
-81x + 729 < 0
when x > 9

f'(x) > 0 when x (is in) [0;9]
f'(x) < 0 when x (is in) [9 ; 13.5[

f raise over [0;9] and decreases on [9 ; 13.5[. So the maximum is fulfilled for x=9
Max = 81* V(3) / 4

Done !


PS : sorry for my bad english and my french calculus fashion