A tank has 40 gallons of water in it at time \( t=0 \) minutes. Water begins to be pumped into the tank at time \( t=0 \). A different pipe is draining water from the tank starting at \( t=0 \). Water is being removed from the tank at a rate modeled by the function \( S(t)=3(1.09)^{t} \), where \( S(t) \) is measured in gallons per minute. Water is being pumped into a tank at a rate modeled by the function \( K(t)=-10(0.9)^{t}+12 \), where \( K(t) \) is measured in gallons per minute. What is the maximum amount of water in the tank from \( t=0 \) to \( t=21 \) ? You may use a calculator and round to the nearest thousandth.

We begin by letting \( W(t) \) represent the amount of water (in gallons) in the tank at time \( t \) (in minutes). We are given that at \( t = 0 \), \[ W(0) = 40. \] Water is being pumped into the tank at a rate modeled by \[ K(t) = -10(0.9)^t + 12, \] and water is drained at a rate modeled by \[ S(t) = 3(1.09)^t. \] Thus, the net rate of change of water in the tank is \[ W'(t) = K(t) - S(t) = -10(0.9)^t + 12 - 3(1.09)^t. \] Notice that \( W'(t) \) represents the instantaneous rate of change of the water volume. To determine the maximum amount of water in the tank over the interval \( t \in [0, 21] \), we must consider where \( W(t) \) reaches its maximum value. In many such problems, we check for critical points by setting \[ W'(t)=0. \] However, let us evaluate \( W'(t) \) at \( t = 0 \): \[ W'(0) = -10(0.9)^0 + 12 - 3(1.09)^0 = -10 + 12 - 3 = -1. \] Since \[ W'(0) < 0, \] the water in the tank is decreasing at \( t = 0 \). Next, observe the behavior of \( W'(t) \) on the interval: - The term \(-10(0.9)^t\) is negative for every \( t \ge 0 \), and the factor \( (0.9)^t \) is a decreasing function. - The constant \( +12 \) remains unchanged. - The term \(-3(1.09)^t\) is negative, and since \( (1.09)^t \) is increasing with \( t \), this negative term becomes more significant as \( t \) increases. Thus, as \( t \) increases from 0 to 21, the net rate \( W'(t) \) remains negative (and indeed becomes even more negative). This means the water volume \( W(t) \) is decreasing continuously throughout the interval \( t \in [0, 21] \). Since the water volume is highest at the very beginning of the time period, the maximum amount of water in the tank occurs at \( t = 0 \) and is given by \[ W(0) = 40 \text{ gallons}. \] Therefore, the maximum amount of water in the tank from \( t = 0 \) to \( t = 21 \) is \[ \boxed{40 \text{ gallons}}. \]