A side-length problem

The area of the triangle is given by: $A = \frac{1}{2}(7)(8)\sin{\theta}$ , where $\theta$ is the angle between 7- and 8-unit sides. $A$ is maximum when $\sin{\theta} = 1$ $\Longrightarrow \theta = 90^\circ$ and the third side $c = \sqrt{7^2+8^2} = \boxed{\sqrt{113}}$

The maximum possible area for a triangle with two given sides comes when those sides form the two legs of a right triangle.

So the length of the third side is $\sqrt{7^2 + 8^2}$ = $\sqrt{113}$