1. **Problem statement:** We have the function $$f(t) = \frac{1}{20}t^3 - \frac{9}{10}t^2 + \frac{77}{20}t$$ which models the rate of change of water volume in a tank (in cubic meters per day) over time $$t$$ in days, for $$0 \leq t \leq 12$$. We want to find the time $$t$$ when the inflow rate is the greatest. 2. **Understanding the problem:** The inflow rate is given by $$f(t)$$. To find when this rate is greatest, we need to find the maximum of $$f(t)$$ on the interval $$[0,12]$$. 3. **Method:** To find maxima, we use the first derivative test. We find $$f'(t)$$, set it to zero to find critical points, then use the second derivative or test values to determine maxima. 4. **Calculate the first derivative:** $$ f'(t) = \frac{d}{dt} \left( \frac{1}{20}t^3 - \frac{9}{10}t^2 + \frac{77}{20}t \right) = \frac{3}{20}t^2 - \frac{18}{10}t + \frac{77}{20} $$ 5. **Simplify coefficients:** $$ f'(t) = \frac{3}{20}t^2 - \frac{9}{5}t + \frac{77}{20} $$ 6. **Set derivative to zero to find critical points:** $$ \frac{3}{20}t^2 - \frac{9}{5}t + \frac{77}{20} = 0 $$ Multiply both sides by 20 to clear denominators: $$ 3t^2 - 36t + 77 = 0 $$ 7. **Solve quadratic equation:** Use the quadratic formula $$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ with $$a=3$$, $$b=-36$$, $$c=77$$: $$ \Delta = (-36)^2 - 4 \cdot 3 \cdot 77 = 1296 - 924 = 372 $$ $$ \sqrt{372} = \sqrt{4 \cdot 93} = 2\sqrt{93} \approx 19.235 $$ So, $$ t = \frac{36 \pm 19.235}{6} $$ Calculate both roots: $$ t_1 = \frac{36 - 19.235}{6} = \frac{16.765}{6} \approx 2.794 $$ $$ t_2 = \frac{36 + 19.235}{6} = \frac{55.235}{6} \approx 9.206 $$ 8. **Determine which critical points are maxima:** Calculate the second derivative: $$ f''(t) = \frac{d}{dt} f'(t) = \frac{d}{dt} \left( \frac{3}{20}t^2 - \frac{9}{5}t + \frac{77}{20} \right) = \frac{3}{10}t - \frac{9}{5} $$ Evaluate $$f''(t)$$ at each critical point: - At $$t_1 \approx 2.794$$: $$ f''(2.794) = \frac{3}{10} \times 2.794 - \frac{9}{5} = 0.838 - 1.8 = -0.962 < 0 $$ Negative second derivative means local maximum. - At $$t_2 \approx 9.206$$: $$ f''(9.206) = \frac{3}{10} \times 9.206 - \frac{9}{5} = 2.762 - 1.8 = 0.962 > 0 $$ Positive second derivative means local minimum. 9. **Check endpoints:** Evaluate $$f(t)$$ at $$t=0$$ and $$t=12$$: $$ f(0) = 0 $$ $$ f(12) = \frac{1}{20} \times 12^3 - \frac{9}{10} \times 12^2 + \frac{77}{20} \times 12 = \frac{1728}{20} - \frac{1296}{10} + \frac{924}{20} = 86.4 - 129.6 + 46.2 = 3.0 $$ 10. **Evaluate $$f(t)$$ at critical points:** $$ f(2.794) = \frac{1}{20} (2.794)^3 - \frac{9}{10} (2.794)^2 + \frac{77}{20} (2.794) \approx \frac{1}{20} (21.8) - 0.9 (7.81) + 3.85 (2.794) = 1.09 - 7.03 + 10.76 = 4.82 $$ $$ f(9.206) = \frac{1}{20} (9.206)^3 - \frac{9}{10} (9.206)^2 + \frac{77}{20} (9.206) \approx \frac{1}{20} (780.9) - 0.9 (84.77) + 3.85 (9.206) = 39.05 - 76.29 + 35.44 = -1.8 $$ 11. **Conclusion:** The maximum inflow rate occurs at $$t \approx 2.79$$ days with $$f(2.79) \approx 4.82$$ cubic meters per day. **Final answer:** The greatest inflow rate is expected at approximately $$t = 2.79$$ days.