下图显示了两个可导函数 $y=f(x)$ 和 $y=g(x)$ 的图像。在 $x = c$ 处，两个函数之间的距离达到最大值。证明 $f'(c)=g'(c)$，其中 $f'(x)$ 和 $g'(x)$ 分别是 $f(x)$ 和 $g(x)$ 的导数。利用二阶导数，清晰地解释为什么该距离是最大值。 两名有着不同策略和能力的跑者 Andy 和 Ben 参加了一场 10000 米的比赛。Andy 和 Ben 在 $x$ 分钟内跑过的距离（以米为单位）分别由以下定义的函数 $A(x)$ 和 $B(x)$ 给出。 $A(x)=\begin{cases} -0.16x^3+12x^2, & 0\le x\le 50 \\ 10000, & x>50 \end{cases}$ $B(x)=5000\log_3(x+9)-10000,\quad 0\le x\le 72.$ 在同一图表上绘制 $0\le x\le 72$ 范围内 $y=A(x)$ 和 $y=B(x)$ 的图像草图。 在 $x = c$ 时，Andy 和 Ben 之间的距离最远。利用 (a) 的结论或通过其他方法，证明 $x = c$ 满足方程 $P x^2 + Q x = \dfrac{R}{x+9},$ 其中 $P, Q$ 和 $R$ 是待确定的常数，并求出 $c$ 的值。 由此或通过其他方法，求出 Andy 和 Ben 之间的最远距离，并说明此时谁处于领先位置。

1. Define the distance function and apply the condition for a maximum to show $f'(c) = g'(c)$.

Let the vertical distance between the two curves be $D(x)$.
From the diagram, $f(x)$ is above $g(x)$ around $x = c$, so:
$D(x) = f(x) - g(x)$
For the distance to be a maximum at $x = c$, the first derivative of $D(x)$ must be zero at this point:
$D'(x) = f'(x) - g'(x)$
$D'(c) = f'(c) - g'(c) = 0$
$\therefore f'(c) = g'(c) \quad \text{(shown)}$

2. Find the second derivative of the distance function.

To confirm a maximum, we use the second derivative test. The second derivative of the distance function is:
$D''(x) = f''(x) - g''(x)$

3. Use the visual concavity of the curves at $x=c$ to determine the sign of $D''(c)$.

From the diagram, at $x = c$, the curve $y = f(x)$ is "hump-shaped" (concave downwards), which means $f''(c) < 0$.
The curve $y = g(x)$ is "trough-shaped" (concave upwards) at $x = c$, which means $g''(c) > 0$.
Substituting these signs into our second derivative expression evaluated at $x=c$:
$D''(c) = f''(c) - g''(c)$
Since $f''(c)$ is negative and we are subtracting a positive $g''(c)$, we get:
$D''(c) < 0$
Because $D''(c) < 0$, the second derivative test confirms that the distance $D(x)$ is a local maximum at $x = c$.

1. Identify the key coordinates and endpoints of both functions.

For $y = A(x)$:
At $x = 0$, $A(0) = 0$.
At $x = 50$, $A(50) = -0.16(50)^3 + 12(50)^2 = -20000 + 30000 = 10000$.
For $50 < x \le 72$, $A(x) = 10000$.

For $y = B(x)$:
At $x = 0$, $B(0) = 5000\log_3(9) - 10000 = 5000(2) - 10000 = 0$.
At $x = 72$, $B(72) = 5000\log_3(81) - 10000 = 5000(4) - 10000 = 10000$.

Both curves start at $(0,0)$ and end at $(72, 10000)$.

2. Determine the shape and relative position of the curves using a graphing calculator.

Using a GC:

• $y = A(x)$ increases as an S-shaped cubic curve from $x=0$ to $x=50$ with a gradient of $0$ at $(0,0)$ and $(50, 10000)$, then remains a horizontal line up to $x=72$.
• $y = B(x)$ is a logarithmic curve that is always increasing and concave downwards.
• $B(x)$ initially rises faster than $A(x)$ near $x=0$, passing above $A(x)$. It crosses $A(x)$ near $x=32$ to go below $A(x)$, remaining below until they meet again at the end point $(72, 10000)$.

3. Sketch the graphs on a single diagram incorporating all key features.

Draw the axes for $0 \le x \le 72$ and $0 \le y \le 10000$.
Sketch both curves $y=A(x)$ and $y=B(x)$ simultaneously, ensuring to highlight the initial tangent of $A(x)$, the single crossing point, the horizontal section of $A(x)$, and the common starting and ending coordinates.

1. Find the derivatives of $A(x)$ and $B(x)$.

For $0 \le x \le 50$:
$A'(x) = 3(-0.16)x^2 + 2(12)x = -0.48x^2 + 24x$

$B(x) = 5000 \log_3(x+9) - 10000$
Using the change of base rule $\log_3(u) = \frac{\ln u}{\ln 3}$, the derivative is:
$B'(x) = 5000 \left( \frac{1}{(x+9)\ln 3} \right) = \frac{5000}{(x+9)\ln 3}$

2. Equate the derivatives and determine the constants $P$, $Q$, and $R$.

Using the result from (a), the distance is at a turning point (and potentially a maximum) when $A'(c) = B'(c)$.

$-0.48c^2 + 24c = \frac{5000}{(c+9)\ln 3}$

Comparing this to the given form $P c^2 + Q c = \frac{R}{c+9}$, we can deduce the constants:
$P = -0.48$
$Q = 24$
$R = \frac{5000}{\ln 3}$

3. Solve the equation for $c$ using a graphing calculator.

We need to solve the equation for $0 \le c \le 50$.
Rearranging to make it easier to enter into a numerical solver:
$(-0.48c^2 + 24c)(c+9) - \frac{5000}{\ln 3} = 0$
$-0.48c^3 + 19.68c^2 + 216c - \frac{5000}{\ln 3} = 0$

Using a graphing calculator to find the roots in the valid range:
$c \approx 11.906 \quad \text{or} \quad c \approx 46.295$

Rounding to 3 significant figures:
$c = 11.9 \text{ or } 46.3$

1. State or solve the equation from (ii) to find the critical times $c$.

From (ii), setting $A'(x) = B'(x)$ to find the points of maximum/minimum distance gives:
$-0.48x^2 + 24x = \frac{5000}{(x+9)\ln 3}$
Using a graphing calculator, we find the roots within the domain $0 \le x \le 50$:
$c_1 \approx 11.906, \quad c_2 \approx 46.295$

2. Evaluate the distance between Andy and Ben at these critical times.

At $c_1 \approx 11.906$:
$A(11.906) = -0.16(11.906)^3 + 12(11.906)^2 \approx 1431.0 \text{ m}$
$B(11.906) = 5000\log_3(11.906+9) - 10000 \approx 3835.8 \text{ m}$
The distance between them is $|1431.0 - 3835.8| \approx 2405 \text{ m}$.

At $c_2 \approx 46.295$:
$A(46.295) = -0.16(46.295)^3 + 12(46.295)^2 \approx 9843.7 \text{ m}$
$B(46.295) = 5000\log_3(46.295+9) - 10000 \approx 8262.5 \text{ m}$
The distance between them is $|9843.7 - 8262.5| \approx 1581 \text{ m}$.

3. Check boundary conditions and conclude the furthest distance and leader.

Checking the boundary when Andy finishes at $x=50$:
$A(50) = 10000 \text{ m}$
$B(50) = 5000\log_3(59) - 10000 \approx 8555 \text{ m}$
The distance is $1445 \text{ m}$.

Comparing these values, the absolute maximum distance is $2405 \text{ m}$ (or $2400 \text{ m}$ to 3 s.f.).
At this time ($x \approx 11.906$), Ben's distance ($3835.8 \text{ m}$) is greater than Andy's ($1431.0 \text{ m}$), so Ben was leading.