Hello Firedawn,
Let's first draw a diagram of the path of the pipeline. All distances are in kilometers.
View attachment 1754
The powerhouse is at $\text{P}$, and the island is at $\text{I}$. The path of the pipeline is drawn in red. Let $C$ be the cost to lay the pipeline over land. The total cost is the cost per unit length time the total length, hence we may express the total cost as a function of $x$ as follows:
$$C(x)=C(13-x)+\frac{7}{5}C\sqrt{x^2+5^2}$$
Differentiating with respect to $x$ and equating the result to zero, we obtain:
$$C'(x)=-C+\frac{7}{5}C\frac{x}{\sqrt{x^2+5^2}}=0$$
Multiply through by $$\frac{5\sqrt{x^2+5^2}}{C}$$
$$-5\sqrt{x^2+5^2}+7x=0$$
$$7x=5\sqrt{x^2+5^2}$$
Square both sides:
$$49x^2=25x^2+625$$
$$x^2=\frac{625}{24}$$
Take the positive root:
$$x=\frac{25}{12}\sqrt{6}\approx5.10310363079829$$
To determine the nature of the extremum associated with this critical value, we may use the second derivative test:
$$C'(x)=-C+\frac{7}{5}C\frac{x}{\sqrt{x^2+5^2}}=0$$
$$C''(x)=0+\frac{7}{5}C\frac{\sqrt{x^2+5^2}(1)-x\left(\dfrac{x}{\sqrt{x^2+5^2}} \right)}{\left(\sqrt{x^2+5^2} \right)^2}=\frac{35C}{\left(x^2+5^2 \right)^{\frac{3}{2}}}$$
We see that for all real $x$ the second derivative is positive, hence our critical value is at the global minimum.
