1. **Problem:** Given the number of cars sold after $t$ days as $$N(t) = 4000 + 45t^2 - t^3$$ for $1 \leq t \leq 48$, find the day when the maximum rate of growth of sales occurs. 2. **Formula and rules:** The rate of growth of sales is the first derivative $$N'(t)$$, and the maximum rate of growth occurs where $$N'(t)$$ is maximized. To find this, we: - Compute $$N'(t)$$. - Find critical points of $$N'(t)$$ by setting its derivative $$N''(t)$$ to zero. - Evaluate $$N'(t)$$ at these points to find the maximum. 3. **Calculate the first derivative:** $$N'(t) = \frac{d}{dt}(4000 + 45t^2 - t^3) = 90t - 3t^2$$ 4. **Calculate the second derivative:** $$N''(t) = \frac{d}{dt}(90t - 3t^2) = 90 - 6t$$ 5. **Find critical points of $$N'(t)$$ by setting $$N''(t) = 0$$:** $$90 - 6t = 0 \implies 6t = 90 \implies t = 15$$ 6. **Check the nature of critical point:** - For $$t < 15$$, $$N''(t) > 0$$ (since 90 - 6t is positive), so $$N'(t)$$ is increasing. - For $$t > 15$$, $$N''(t) < 0$$, so $$N'(t)$$ is decreasing. Thus, $$t=15$$ is a maximum point for $$N'(t)$$. 7. **Conclusion:** The maximum rate of growth of sales occurs on day $$15$$. **Final answer:** (C) on day 15