Minimizing total pipeline cost with land and underwater segments

The underwater distance from $A$ to $C$ is $\sqrt{x^2+16}$ by the Pythagorean theorem, since the river is $4$ miles wide and $x$ is the horizontal offset from $P$. The land distance from $C$ to $B$ is $10-x$. Thus the total cost function is $C(x) = 4\sqrt{x^2+16} + 2(10-x), \quad 0 \le x \le 10.$

Differentiating gives $C'(x) = \dfrac{4x}{\sqrt{x^2+16}} - 2$. Setting $C'(x)=0$ leads to $4x = 2\sqrt{x^2+16}$, so $2x = \sqrt{x^2+16}$, and squaring gives $4x^2 = x^2+16$, so $x^2 = 16/3$, giving $x = 4/\sqrt{3} \approx 2.31$. Checking this critical point against the endpoints $x=0$ and $x=10$ using the second derivative (which is positive, confirming a minimum) or direct cost comparison confirms $x = 4/\sqrt{3}$ minimizes total cost.

Yes. Increasing the underwater cost per mile makes the underwater route relatively more costly, so the optimal strategy shifts toward minimizing underwater distance. This pushes the optimal point C closer to P, reducing the underwater segment length while increasing the land segment. This follows directly from how the cost function balances the two rates in the original setup.