#1**+1 **

[ 55t ] / [ t^2 + 25 ] ≥ 5 multiply both sides by t^2 + 25

55t ≥ 5 [ t^2 + 25] simplify

55t ≥ 5t^2 + `125 divide through by 5

11t ≥ t^2 + 25 subtract 11t from both sides

0 ≥ t^2 - 11t + 25 and we can write

t^2 - 11t + 25 ≤ 0

Let's solve this

t^2 - 11t + 25 = 0

Putting this into the quadratic formula......we have the [approximate] solutions

t ≈ 3.2 hrs and t ≈ 7.8 hrs

Choosing a test value between these two values [ I'll choose 4 ] and putting this into

t^2 - 11t + 25 ≤ 0 → 4^2 - 11(4) + 25 = 16 - 44 + 25 = -3

So....since 4 makes the inequality true.....the interval between 3.2 hrs and 7.8 hrs is the correct solution.....i.e, the concentration will be ≥ 5 between these two times

Here's the graph, substituting x for t : https://www.desmos.com/calculator/cjgoqwvrj5

Notice that [ 55t ] / [ t^2 + 25 ] is ≤ 5 between these two times

CPhill
Oct 11, 2017