
Yefim S. answered 11/02/21
Math Tutor with Experience
Q'(t) = 16.78·0.15cos(0.15t - 1.25) = 0; cos(0.15t - 1.25) = 0.15t - 1.25 = π/2; t = (π/2 + 1/25)/0.15 = 18.8 sec
Q''(t) = - 16.75·0.152sin(0.15t -1.25); Q''(18.8) = - 16.75·0.152sin(0.15 ·18.8 - 1.25) = - 0.377 < 0.
So, at t = 18.8 sec we have maximum leak