For example, as shown in Figure $24-2$, a tunnel is designed for four lanes in both directions, with a total lane width of 22 meters, and the vehicle height limit is 4.5 meters. The total length of the tunnel is 25 kilometers, and the arch of the tunnel is approximately a half-ellipse shape.
(1) If the maximum arch height $h$ is 6 meters, what is the arch width $l$ of the tunnel design?
(2) If the maximum arch height $h$ is not less than 6 meters, how should the arch height $h$ and arch width $l$ be designed to minimize the earthwork volume of the half-elliptical tunnel? (The area formula for a half-ellipse is $S=\frac{\pi}{4} l h$, and the volume of a prism is: base area multiplied by height. The results of this problem should be accurate to 0.1 meters. 2003 Shanghai College Entrance Examination Question)
(1) Establish a rectangular coordinate system as shown in the figure, then point $P(11,4.5)$, the equation of the ellipse is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$.
Substitute $b=h=6$ and the coordinates of point $P$ into the equation of the ellipse, we get $a=\frac{44}{7} \sqrt{7}, l=2 a=\frac{88}{7} \sqrt{7} \approx 33.3$. That is, the width of the tunnel arch is approximately 33.3 meters.
(2) From the equation of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, we have $\frac{11^{2}}{a^{2}}+\frac{4.5^{2}}{b^{2}}=1$. And $\frac{11^{2}}{a^{2}}+\frac{4.5^{2}}{b^{2}} \geqslant \frac{2 \times 11 \times 4.5}{a b}, a b \geqslant 99$, $l=2 a, h=b . S=\frac{\pi}{4} l h=\frac{\pi}{2} a b \geqslant \frac{99}{2} \pi$. When the equality holds, $\frac{11^{2}}{a^{2}}=\frac{4.5^{2}}{b^{2}}=\frac{1}{2}, a=11 \sqrt{2}, b=$ $\frac{9}{2} \sqrt{2}$, at this time, $l=2 a=22 \sqrt{2} \approx 31.1, h=b \approx 6.4$. Therefore, when the arch height is approximately 6.4 meters and the arch width is approximately 31.1 meters, the earthwork volume is minimized.
!Note This problem is related to ellipses in analytic geometry, and the basic inequality also plays a significant role in part (2).