The Bretschneider's formula for a general quadrilateral is $\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd*\cos^2\bigg(\frac{\alpha+\gamma}{2}\bigg)}$ This is maximized if the quadrilateral is inscribed in a circle. We see that $1^2+8^2=4^2+7^2=65.$ Thus, it is inscribed in a circle with radius $\frac{\sqrt{65}}{2}.$ We can break it up into $2$ right triangles with areas $4$ and $14.$ Thus our answer is $4+14=\boxed{18}$