1. **State the problem:** We have a drug amount function $$C(t) = -t^3 + 4.5t^2 + 54t$$ representing the amount of drug in the bloodstream after $$t$$ hours. 2. **Find the rate of change of the drug amount:** The rate of change is the derivative $$C'(t)$$ of the function $$C(t)$$. 3. **Formula for derivative:** Using power rule $$\frac{d}{dt}[t^n] = nt^{n-1}$$, we differentiate each term: $$C'(t) = \frac{d}{dt}[-t^3] + \frac{d}{dt}[4.5t^2] + \frac{d}{dt}[54t]$$ $$= -3t^2 + 9t + 54$$ 4. **Evaluate rate of change after 4 hours:** Substitute $$t=4$$ into $$C'(t)$$: $$C'(4) = -3(4)^2 + 9(4) + 54 = -3(16) + 36 + 54 = -48 + 36 + 54$$ $$= 42$$ So, the drug amount is increasing at a rate of 42 units per hour after 4 hours. 5. **Find maximum drug amount over first 9 hours:** Maximum occurs where $$C'(t) = 0$$ (critical points). Solve: $$-3t^2 + 9t + 54 = 0$$ Divide both sides by -3: $$\cancel{-3}t^2 + \cancel{3}t + \cancel{18} = 0 \Rightarrow t^2 - 3t - 18 = 0$$ Use quadratic formula: $$t = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(-18)}}{2} = \frac{3 \pm \sqrt{9 + 72}}{2} = \frac{3 \pm \sqrt{81}}{2}$$ $$= \frac{3 \pm 9}{2}$$ So, $$t=6$$ or $$t=-3$$ (discard negative time). 6. **Check if $$t=6$$ is max:** Find second derivative: $$C''(t) = \frac{d}{dt}C'(t) = \frac{d}{dt}(-3t^2 + 9t + 54) = -6t + 9$$ Evaluate at $$t=6$$: $$C''(6) = -6(6) + 9 = -36 + 9 = -27 < 0$$ Since $$C''(6) < 0$$, $$t=6$$ is a local maximum. 7. **Calculate maximum drug amount:** $$C(6) = -(6)^3 + 4.5(6)^2 + 54(6) = -216 + 4.5(36) + 324 = -216 + 162 + 324 = 270$$ 8. **Show drug amount decreasing at $$t=7$$:** Evaluate $$C'(7)$$: $$C'(7) = -3(7)^2 + 9(7) + 54 = -3(49) + 63 + 54 = -147 + 63 + 54 = -30$$ Since $$C'(7) < 0$$, the drug amount is decreasing at 7 hours. **Final answers:** (i) $$C'(t) = -3t^2 + 9t + 54$$ (ii) Rate at 4 hours: 42 units/hour (iii) Maximum amount: 270 units at 6 hours (iv) Drug amount decreasing at 7 hours because $$C'(7) < 0$$