For largest triangle
$x = y = z$
 

 

$A_{max} = \frac{1}{2}(6^2) \sin 60^\circ$

$A_{max} = 15.588 ~ \text{cm}^2$

Perimeter of the triangle
$x + y + z = 18$

$z = 18 - x - y$   ←   Eq. (1)
 

Semi-perimeter of the triangle
$s = \frac{1}{2}(18)$

$s = 9 ~ \text{cm}$
 

 

Area of the triangle by Heron's formula
$A = \sqrt{s(s - x)(s - y)(s - z)}$   ←   Eq. (2)

$A = \sqrt{9(9 - x)(9 - y)\left[ 9 - (18 - x - y) \right]}$

$A = \sqrt{9(9 - x)(9 - y)(x + y - 9)}$
 

For maximum possible area of the triangle

$\dfrac{\partial A}{\partial x} = \dfrac{9(9 - y) \cdot \left[ (9 - x)(1) + (x + y - 9)(-1) \right]}{2\sqrt{9(9 - x)(9 - y)(x + y - 9)}} = 0$

$9 - x = x + y - 9$

$y = 18 - 2x$   ←   Eq. (3)
 

$\dfrac{\partial A}{\partial y} = \dfrac{9(9 - x) \cdot \left[ (9 - y)(1) + (x + y - 9)(-1) \right]}{2\sqrt{9(9 - x)(9 - y)(x + y - 9)}} = 0$

$9 - y = x + y - 9$

$x = 18 - 2y$   ←   Eq. (4)

 

Substitute $y = 18 - 2x$ from Eq. (3) to Eq. (4)
$x = 18 - 2(18 - 2x)$

$3x = 18$

$x = 6 ~ \text{cm}$
 

Substitute $x = 6$ to Eq. (3)
$y = 18 - 2(6)$

$y = 6 ~ \text{cm}$
 

Substitute $x = 6$ and $y = 6$ to Eq. (1)
$z = 18 - 6 - 6$

$z = 6 ~ \text{cm}$
 

Hence, from Eq. (2)
$A_{max} = \sqrt{9(9 - 6)(9 - 6)(9 - 6)}$

$A_{max} = 15.588 ~ \text{cm}^2$