The first thing to understand is that when you write down the cost function $C$, you also must be mindful of the domain for which that cost function is meaningful. That is to say, if $C(x)$ represents the number of hundreds of dollars when the pivot point $P$ is situated $x$ kilometers from $B$, then clearly the diagram makes sense only when $0 \le x \le k$, and this is the domain.
Moreover, we observe that the expression $$C(x) = 125\sqrt{x^2 + 5^2} + 100(k-x)$$ would assign a negative cost to the second term in the cost function when $x > k > 0$. This ideally would be rectified by writing $$C(x) = 125 \sqrt{x^2 + 5^2} + 100|k-x|$$ instead.
Consequently, when you solved for the critical points satisfying $C'(x) = 0$ and found that the only critical point occurs at $x = 20/3$, and that this value is independent of $k$, what you have essentially deduced are two facts:
- When $k < 20/3$, the cost associated with placing the pivot $P$ at $x = 20/3$ is not valid.
- The cost function does not attain its minimum at a critical point of $C$, but instead must do so at an endpoint of the interval $[0,k]$.
Specifically, then, when $k < 20/3$, it is cheapest to simply lay the line directly from $A$ to $F$. No cost savings are achieved by choosing an intermediate point $P$ to run the line along land, then from $P$ to $A$, because the cost savings of running a portion of the cable along land is not adequately offset by the resulting increase in the total distance of cable needed. Only when $k$ is sufficiently large, and thus the hypotenuse $\overline{AF}$ sufficiently long, does it make sense to reduce the distance of cable under water.
As a follow-up exercise, then, in the general case where the cost of installing underwater cable is $\alpha$ per kilometer and the cost on land is $\beta$ per kilometer, with $\alpha > \beta$, what is the largest $k > 0$ in terms of $\alpha$ and $\beta$ for which the most efficient path is the hypotenuse $\overline{AF}$?
Let's start with the components of the velocity of the boat $\vec{V}$
$$ \begin{align} V_x&=c \cos \theta \\ V_y&=r + c \sin \theta \text{ }\end{align} $$
where $\theta$ is the steering angle and $r(x)$ is the river speed you called $v(x)$. Clearly, we have that the time it takes to cross the river is given by
$$ T = \int \text{d}t = \int \frac{\text{d}x}{\|V_x\|} = \int \frac{\text{d}x}{ c \cos \theta } = \int \frac{1}{c}\sec \theta \: \text{d}x \text{ .}$$
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By the chain rule, if we say that $y=y(x)$ then
$$ \dfrac{\text{d}y}{\text{d}t} = \dfrac{\text{d}y}{\text{d}x} \: \dfrac{\text{d}x}{\text{d}t} $$
and so we get
$$\begin{align} &\dfrac{\text{d}y}{\text{d}x} = \frac{\dot{y}}{\dot{x}} = \frac{V_y}{V_x} = \frac{r + c \sin \theta}{c \cos \theta}\\\\ \implies &\dfrac{\text{d}y}{\text{d}x} =\frac{r}{c}\sec \theta \: + \tan \theta \\\\ \implies &\dfrac{\text{d}y}{\text{d}x} = \frac{r}{c}\sec \theta \: + \sqrt{\sec^2 \theta - 1}\end{align}$$
just by using a neat trig identity. This is now a quadratic for $\sec \theta$ so we rearrange as
$$ \bigg( \frac{r^2}{c^2} - 1 \bigg)\sec^2 \theta \: - \frac{2ry'}{c}\sec{\theta} \: + (y')^2 + 1 = 0 $$
where $\dfrac{dy}{dx}=y'$. We use quadratic formula to obtain
$$ \sec{\theta} = \frac{\frac{2ry'}{c} \pm \sqrt{ 4(y'^2 - \frac{r^2}{c^2} + 1) }}{ 2\bigg(\frac{r^2}{c^2} - 1\bigg) } \text{ .} $$
I'm simplifying a tad because I'm too lazy to write it all out. What we get is that the $b^2$ term in $b^2 - 4ac$ cancels with part of the $4ac$.
We can rewrite the denominator as
$$ 2\bigg(\frac{r^2}{c^2} - 1\bigg) = 2\bigg(\frac{r^2 - c^2}{c^2}\bigg) = -2\bigg(\frac{c^2 - r^2}{c^2}\bigg) $$
which gives us that
$$ \begin{align} &\frac{1}{c}\sec{\theta} = \frac{-ry' \mp c\sqrt{ y'^2 - \frac{r^2}{c^2} + 1 }}{ \bigg(c^2 - r^2\bigg) } \\\\ \implies
&\frac{1}{c}\sec{\theta} = \frac{-ry' \mp \sqrt{ c^2(y'^2+1) - {r^2} }}{ \bigg(c^2 - r^2\bigg) } \text{ .} \end{align} $$
Take the appropriate root for $\sec\theta$ and substitute it into our functional for $T$.
If I skipped over too much and you can't seem to fill in the details just let me know and I'll make the appropriate edits.
Best Answer
The problem is that the critical point you found is not on the domain! The values for any solution must be $x\in[0,3]$. Since your critical point lies outside the domain, you have to check the endpoints for your solution. In this case the answer should be $x=3$. In other words, he should row all the way!