15. A steel bar with a total length of $14.8 \mathrm{~m}$ is used to make a framework of a rectangular container. If one side of the base of the container is $0.5 \mathrm{~m}$ longer than the other, what should the height be to maximize the volume of the container? Also, find the maximum volume.

Using a steel bar with a total length of $14.8 \mathrm{~m}$ to make a framework of a rectangular container, if one side of the base of the container is $0.5 \mathrm{~m}$ longer than the other, then when the height is what value will the volume of the container be maximized? And find its maximum volume.
15. 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 $\frac{14.8-4 x-4(x+0.5)}{4}=3.2-2 x$. And $3.2-2 x>0, x>0$, thus $x \in(0,1.6)$. Let the volume of the container be $y \mathrm{~m}^{3}$, then $y=x(x+0.5)(3.2-2 x)=-2 x^{3}+2.2 x^{2}+1.6 x$. Taking the derivative, $y'=-6 x^{2}+4.4 x+1.6$, solving for $y'=0$ gives $x=1$ or $x=-\frac{4}{15}$ (discard), thus within the domain, $y'=0$ only at $x=1$. Therefore, when $x=1$, $y$ has a maximum value $y_{\text{max}}=1 \times 1.5 \times 1.2=1.8$, at this time the height is $1.2 \mathrm{~m}$.