(b) A $10 k L$ tank of water provides water for a small village. Water is drunk from the tank at a rate of $1000 \mathrm{~L}$ per day, and the water is replenished at the same rate. Unfortunately, the filter/purifier for the intake water is malfunctioning. Consequentially the water that is replenishing the tank is contaminated; it contains $0.00125 \mathrm{mg} / \mathrm{L}$ of heavy metals. Suppose that when the contamination was discovered the tank was $3 / 4$ full and contained $a_{s} m g$ of heavy metals. (i) Find a formula, $A(t)$, for the amount of heavy metals in the tank at time $t$. Hint: make sure your units are consistent for all quantities. We suggest $\mathrm{mg} / \mathrm{kL}$ If the concentration of heavy metals in the water is beldw $0.001 \mathrm{mg} / \mathrm{L}$ then the water is still safe to drink. You may assume that the heavy metals in the tank are well mixed (and thus evenly distributed) in the tank. The maintenance company has been notified, and have indicated that the they'll be on premises to service the filter/purifier in 7 days (and please make sure somebody is on premises during that entire day). However, we know from experience that they can be anywhere up to 3 days late. (ii) If the initial amount of heavy metals in the tank was $2 m g$, will the concentration reach dangerous levels before the company services the tank? (Yes, no, or maybe?). Justify your answer. (iii) What is the largest initial amount of heavy metals that will definitely not reach dangerous levels before the company services the tank?

Step 1 ：Let the amount of heavy metals in the tank at time t be A(t) in mg/kL. The tank is 3/4 full, so it contains 7.5 kL of water. The initial amount of heavy metals is a_s mg.

Step 2 ：At time t, the tank has 7.5 - 1000t/1000 kL of water left and 1000t/1000 kL of contaminated water added. So, the total volume of water in the tank at time t is 7.5 kL.

Step 3 ：The amount of heavy metals in the tank at time t is the sum of the initial amount a_s and the amount added by the contaminated water, which is 0.00125 * 1000t mg. So, A(t) = a_s + 0.00125 * 1000t.

Step 4 ：Convert A(t) to mg/kL by dividing by the total volume of water in the tank: A(t) = (a_s + 0.00125 * 1000t) / 7.5.

Step 5 ：\(\boxed{A(t) = \frac{a_s + 1.25t}{7.5}}\)

Step 6 ：For part (ii), we need to find if the concentration reaches 0.001 mg/L before the company services the tank. The latest they can arrive is in 10 days. So, we need to check if A(10) is less than 0.001 mg/L.

Step 7 ：Substitute a_s = 2 mg and t = 10 days into the formula: A(10) = (2 + 1.25 * 10) / 7.5.

Step 8 ：Calculate A(10): A(10) = (2 + 12.5) / 7.5 = 14.5 / 7.5 = 1.933 mg/kL.

Step 9 ：Since 1.933 mg/kL is greater than 0.001 mg/L, the concentration will reach dangerous levels before the company services the tank. So, the answer is \(\boxed{\text{Yes}}\).

Step 10 ：For part (iii), we need to find the largest initial amount of heavy metals that will not reach dangerous levels before the company services the tank in 7 days.

Step 11 ：Set A(7) equal to 0.001 mg/L and solve for a_s: 0.001 = (a_s + 1.25 * 7) / 7.5.

Step 12 ：Cross multiply and solve for a_s: a_s = 0.001 * 7.5 - 1.25 * 7.

Step 13 ：Calculate a_s: a_s = 0.0075 - 8.75 = -8.7425 mg.

Step 14 ：Since a negative amount of heavy metals is not possible, the largest initial amount of heavy metals that will definitely not reach dangerous levels before the company services the tank is \(\boxed{0}\).