Answer
See below
Work Step by Step
From part c, we found $$A_1=\frac{391t^2+0.112}{0.218t^4+0.991t^2+1}\\A_2=\frac{391(t-1)^2+0.112}{0.218(t-1)^4+0.991(t-1)^2+1}\\A=\frac{391t^2+0.112}{0.218t^4+0.991t^2+1}+\frac{391(t-1)^2+0.112}{0.218(t-1)^4+0.991(t-1)^2+1}$$
The maximum of $A$ occurs at about $t=2.4$ with about $203$ milligrams.