9. (2000 National College Entrance Examination for Two Provinces and One Municipality) A rectangular container frame is made from a steel strip with a total length of 14.8 meters. If one side of the container's base is 0.5 meters longer than the other, what should the height be to maximize the volume of the container? Also, find its maximum volume.

Use the total length of 14.8 meters of steel strip to make a frame of a rectangular container. If one side of the base of the container is 0.5 meters longer than the other, what should the height be to maximize the volume of the container? And find its maximum volume.
9. Let the length of the shorter side of the container's base be $x \mathrm{~m}$, then the length of the other side is $(x+0.5) \mathrm{m}$, and the height is
$$
\frac{14.8-4 x-4(x+0.5)}{4}=3.2-2 x \text {. }
$$

From $3.2-2 x>0$ and $x>0$, we get $0<x<1.6$.
The volume of the container is
$$
y=x(x+0.5)(3.2-2 x)=-2 x^3+2.2 x^2+1.6 x \quad (0<x<1.6) \text {. }
$$

Differentiating $y$ with respect to $x$, we get
$$
y'=-6 x^2+4.4 x+1.6 \text {. }
$$

Setting $y'=0$, we solve to get $x_1=-0.2, x_2=1.333$.
Since $x_1$ is not in the interval $(0, 1.6)$, we discard it.
When $0<x<1$, $y'>0$; when $1<x<1.6$, $y'<0$,
so, $x=1$ is the point of maximum value for the function $y$, the maximum value is $1 \times 1.5 \times 1.2=1.8$.
When $x=0$ or $x=1.6$, $y=0$,
thus, when $x=1$, the function $y$ has a maximum value of 1.8, at this time the height of the container is $1.2 \mathrm{~m}$.