Max. and Min.

Let V cm³ be the volume of the cube at t min.

V = (10 + 5t) x (10 + 2t) x (10 - t)
V = 1000 + 600t + 30t² - 10t³

dV/dt = 600 + 60t - 30t²
dV/dt = -30(t² - 2t - 20)
dV/dt = -30 [t - (1 + √21)] [t - (1 - √21)]
d²V/dt² = 60 - 60t

When t = 1 + √21 :
dV/dt = 0
d²V/dt² = 60 - 60(1 + √2) = -60√2< 0
V = 1000 + 600(1 + √21) + 30(1 + √21)² - 10(1 + √21)³
V = 1620 + 420√21
When t = 1 + √21, max. V = 1620 + 420√21

Since V ≥ 0
When t = 0, min. V = 0

(1)
Max. volume = (1620 + 420√21) cm³
Min. volume = 0 cm³

(2)
Dimensions at max. volume = (15 + 5√21) cm x (12 + 2√21) cm x (9 - √21) cm
Dimensions at min. volume = 60 cm x 30 cm x 0 cm

(3)
Time when volume is at max. = (1 + √21) min
Time when volume is at min. = 10 min

你喺咪講緊 - 1 cm./min (負一cm/min)，咁最低係幾多﹖會唔會去到零﹖

唔會嗎﹖ 咁你只有九分鐘時間，最後大小係50cm x 28cm x 1cm

咁就1,2,3 都答哂你﹗