1. **State the problem:** We have a fish population modeled by the function $$p(t) = 15(t^2 + 30)(t + 8)$$ where $t$ is time in years. We need to find the rate of change of the fish population when there are 5000 fish in the lake. 2. **Find the derivative $p'(t)$ to get the rate of change:** Use the product rule for derivatives: If $$p(t) = f(t)g(t)$$ then $$p'(t) = f'(t)g(t) + f(t)g'(t)$$. Here, let $$f(t) = 15(t^2 + 30)$$ and $$g(t) = (t + 8)$$. 3. **Calculate derivatives:** $$f'(t) = 15 \cdot 2t = 30t$$ $$g'(t) = 1$$ 4. **Apply product rule:** $$p'(t) = 30t(t + 8) + 15(t^2 + 30)(1)$$ 5. **Simplify:** $$p'(t) = 30t^2 + 240t + 15t^2 + 450 = (30t^2 + 15t^2) + 240t + 450 = 45t^2 + 240t + 450$$ 6. **Find $t$ when $p(t) = 5000$:** $$p(t) = 15(t^2 + 30)(t + 8) = 5000$$ Divide both sides by 15: $$\cancel{15}(t^2 + 30)(t + 8) = \cancel{15} \times \frac{5000}{15}$$ $$ (t^2 + 30)(t + 8) = \frac{5000}{15} = 333.33$$ 7. **Expand left side:** $$t^3 + 8t^2 + 30t + 240 = 333.33$$ 8. **Bring all terms to one side:** $$t^3 + 8t^2 + 30t + 240 - 333.33 = 0$$ $$t^3 + 8t^2 + 30t - 93.33 = 0$$ 9. **Solve cubic approximately:** By trial or numerical methods, $t \approx 2$ years. 10. **Calculate rate of change at $t=2$:** $$p'(2) = 45(2)^2 + 240(2) + 450 = 45 \times 4 + 480 + 450 = 180 + 480 + 450 = 1110$$ **Final answer:** The rate of change of the fish population when there are 5000 fish is approximately $$1110$$ fish per year.