Finding Maximum area of the isosceles triangle $A = \{(x , y ,z): 1 \leq x , y, z \leq 6$ and $x , y ,z$ are the sides of a isosceles triangle$\}$
How can I find the maximum area of the triangle whose sides $a , b , c$ , $(a , b , c) \in A$?
Can anyone please give me some hint?
 A: Let us first find out the set of sample points $S = \{(x , y ,z) : 1 \leq x , y , z \leq 6$ and  $ x , y, z $ are the sides of an isosceles triangle $\} $.\
So $S= \{(2,2,1),(2,2,3),(3,3,1),(3,3,2),(3,3,4),(3,3,5),(4,4,1),\\
(4,4,2),(4,4,3),(4,4,5),(4,4,6),(5,5,1),(5,5,2),(5,5,3)\\
(5,5,4)(5,5,6)(6,6,1)(6,6,2)(6,6,3)(6,6,4)(6,6,5)\\
 (1,1,1)(2,2,2)(3,3,3)(4,4,4)(5,5,5)(6,6,6)\}$\
Total number of elements of $S$ is $27$.\
Now we know area of a triangle with sides $a , b ,c$ is\
$A = \sqrt{x(x-a)(x-b)(x-c)}$ where $x = \frac{a+b+c}{2}$\
$ = \frac{1}{4} \sqrt{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}$.
Now we need to find $\max _{(a, b,c)\in S}\frac{1}{4} \sqrt{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}$.\
Let's say $x = (b+c-a) $\ $y = (a+c-b) $\ $z = (a+b-c) $.\
Now as $a , b ,c$ are the sides of a triangle we can say $x>0$ , $y>0$ , $z>0$ .\
So we can say $\frac{x+y}{2} = c$ , $\frac{z+y}{2} = a$ , $\frac{x+z}{2} = b$\
Now we can write \
$ \sqrt xy \sqrt xz \sqrt yz \leq \frac{x+y}{2}  \frac{x+z}{2} \frac{z+y}{2}$  [By applying A.M $\geq$ G>M ]\
$\Rightarrow (b+c-a) (a+c-b) (a+b-c) \leq abc$  \
$\Rightarrow \frac{1}{4} \sqrt{(a+b+c)(b+c-a)(a+c-b)(a+b-c)} \leq \frac{1}{4} \sqrt{ (a+b+c)abc}$\
[We can do so because $(b+c-a)(a+c-b)(a+b-c)  >0$ and $abc >0$ ]\
If $a = b= c$ then the equality occurs.\ .
Now we can say for a particular triangle with sides of lengths $ m , n , p$ and $ m \geq n \geq p$ ,
the area of the triangle with sides of length $m$ will always be greater than the area of the triangle with sides of lengths $ m , n, p$.\
So we can say $A$ will assume it's maximum when $a = b= c = 6$. As all points of $S$ are $(x , y , z)$ where  $1 \leq x , y , z \leq 6$.
$\max _{(a, b,c)\in S}\frac{1}{4} \sqrt{(a+b+c)(b+c-a)(a+c-b)(a+b-c)} = \frac{1}{4} \sqrt{(6+6+6)6.6.6} =9 \sqrt3$.
    So we will have only one point in the set of sample points $S$ , $(6 , 6 , 6)$  ,for which the triangle will be of maximum area.\
the probability that the triangle is of maximum area given that it is an isosceles triangle, is $\frac{1}{27}$.
A: An Isosceles is composed of $2$ right triangles and, for Pythagorean triples, the one with the largest area compared to leg length is $3,4,5$. The area of one such triangle is $3\times2=6$ and putting two of them together by joining the altitude $(4)$ yields a triangle of $5,5,6$ with an area of $12$.
Suppose we tried a larger triangle and reduced it. The area of $21,20,29$ is $21\times10=210$. Dividing by $7$ and joining on altitude $\frac{20}{7}$, we get $\frac{29}{7},\frac{29}{7},6$ with an area of $$\frac{2\times210}{7^2}=8.571428571428571$$
or
$$2\times3\times \frac{10}{7}=8.571428571428571$$
The largest area is with the first example, i.e. $2$ sides of $5$ and a base of $6$.
On the other hand, if the isosceles triangle is also equilateral, you can have side lengths of $6$ and an altitude of $5.196152423$ so the area would be $3\times 5.196152423=15.58845727$.
