The depth of water in a reservoir, DDD metres, was recorded over a 10-day period. The depth at time ttt days, where 0≤t≤100 \le t \le 100≤t≤10, is modeled by the equation:
D=t30(18+8t−t2)+12 D = \frac{\sqrt{t}}{30}(18 + 8t - t^2) + 12 D=30t(18+8t−t2)+12Given that DDD has a stationary value at t=αt = \alphat=α:
Use calculus to show that α\alphaα satisfies the equation
5α2−24α−18=0 5\alpha^2 - 24\alpha - 18 = 0 5α2−24α−18=0Hence find the value of α\alphaα, giving your answer to 3 decimal places.
Use further calculus to prove that DDD is a maximum at this value of α\alphaα.
