6. A city's waterworks, the reservoir originally has 500 tons of water. In one day, while water is being added to the reservoir, water is also being supplied to users from the reservoir. $x(0 \leqslant x \leqslant 24)$ hours of water supplied to users is $80 \sqrt{24 x}$ tons. When 96 tons of water are added to the reservoir per hour, then the minimum amount of water in the reservoir in one day is ( ) tons.
(A) 120
(B) 100
(C) 80
(D) 60
6. (B).

From the given, we have
$$
y=500+96 x-80 \sqrt{24 x}(0 \leqslant x \leqslant 24) \text {. }
$$

Let $\sqrt{24 x}=t(0 \leqslant t \leqslant 24)$. Then
$$
y=4 t^{2}-80 t+500=4(t-10)^{2}+100 \text {. }
$$

Therefore, when $t=10$, the minimum value of $y$ is 100.