在七天内，监测河面相对于河床某固定点的高度。监测开始后 $t$ 天，河面高度 $H$ 米满足

已知 $H$ 在 $t=\alpha$ 处有驻值，

(a) 使用微积分证明 $\alpha$ 满足 $5\alpha^2-18\alpha-20=0$。（5 分）

(b) 由此求 $\alpha$，答案保留三位小数。（1 分）

(c) 使用进一步的微积分证明 $H$ 在该 $\alpha$ 值处取得极大值。（2 分）

将含根号的部分展开为幂函数，逐项求导。令驻值处的一阶导数为零，消去分数指数并整理为关于 $\alpha$ 的二次方程。

The height is $H = \dfrac{1}{20}(20 + 6t - t^2)\sqrt{t} + 17$, i.e.

At the stationary value $t = \alpha$, $\dfrac{\mathrm{d}H}{\mathrm{d}t} = 0$:

Multiply through by $\dfrac{2}{\alpha^{1/2}}$:

Multiply by $\alpha$:

将 $H-17$ 看成多项式因子与 $t^{1/2}$ 的乘积，用乘积法则求导，再令一阶导数为零。

Write $H = \dfrac{1}{20}(20 + 6t - t^2) \cdot t^{1/2} + 17$.

Using the product rule with $u = \dfrac{1}{20}(20 + 6t - t^2)$ and $v = t^{1/2}$:

Setting $\dfrac{\mathrm{d}H}{\mathrm{d}t} = 0$ and multiplying through by $20t^{1/2}$:

At $t = \alpha$: $5\alpha^2 - 18\alpha - 20 = 0$. $\quad\blacksquare$

承接 (a)，用二次公式求两个根，按 $0\leqslant t\leqslant7$ 选择正根。

Using the quadratic formula:

Since $0 \leqslant \alpha \leqslant 7$, take the positive root:

再次求导，并将未舍入的 $\alpha$ 代入。二阶导数为负，因此该驻值是极大值。

The second derivative:

At $t = \alpha \approx 4.491$:

Using the unrounded value $\alpha=(9+\sqrt{181})/5$,

Since $\dfrac{\mathrm{d}^2H}{\mathrm{d}t^2} < 0$ at $t = \alpha$, $H$ has a maximum at this value of $\alpha$. $\quad\blacksquare$

将正确的二阶导数提取负因子，证明余下二次式恒正，于是二阶导数在正的 $t$ 上恒负。

The second derivative can be written as:

For the quadratic $15t^2 - 18t + 20$, the discriminant is:

Since the leading coefficient $15 > 0$ and $\Delta < 0$, we have $15t^2 - 18t + 20 > 0$ for all $t$.

Also, $t^{-3/2} > 0$ for $t > 0$.

Therefore $\dfrac{\mathrm{d}^2H}{\mathrm{d}t^2} < 0$ for all $t \in (0, 7)$, so $H$ has a maximum at $t = \alpha$. $\quad\blacksquare$