An isosceles triangle has integer sides and has perimeter 16. Find the largest possible area of the triangle.

With equal sides $$a$$ and base $$b = 16 - 2a$$, the triangle inequality needs $$b < 2a$$, leaving $$(5,5,6)$$, $$(6,6,4)$$ and $$(7,7,2)$$. Heron's formula with $$s = 8$$ gives areas $$\sqrt{8 \times 3 \times 3 \times 2} = 12$$, $$\sqrt{8 \times 2 \times 2 \times 4} = 8\sqrt{2}$$ and $$\sqrt{8 \times 1 \times 1 \times 6} = 4\sqrt{3}$$. The largest of these is 12.