6. (3 marks) The cost of laying a fibre optic cable underwater is 3 times that of underground. An island is $4 \mathrm{~km}$ from point $S$ on the shore and a broadcast station, $B$, is $14 \mathrm{~km}$ along the shore, perpendicular to S. How should the cable be laid to keep cost to a minimum?

Step 1 ：Define the variable x and the cost function. The cost function is given by the equation \(cost = -x + 3\sqrt{x^2 + 16} + 14\).

Step 2 ：Calculate the derivative of the cost function. The derivative is given by the equation \(cost\_derivative = 3x/\sqrt{x^2 + 16} - 1\).

Step 3 ：Solve the equation where the derivative equals to zero to find the critical points. The critical point is \(\sqrt{2}\).

Step 4 ：Check the second derivative to make sure the solution is a minimum. The second derivative is given by the equation \(second\_derivative = -3x^2/(x^2 + 16)^{3/2} + 3/\sqrt{x^2 + 16}\). The second derivative at the critical point is positive, which means this point is a minimum.

Step 5 ：Calculate the minimum cost by substituting the critical point into the cost function. The minimum cost is \(3\sqrt{18} + 14 - \sqrt{2}\).

Step 6 ：The cable should be laid to the point on the shore that is \(\sqrt{2} \mathrm{~km}\) from point S. The minimum cost is \(\boxed{3\sqrt{18} + 14 - \sqrt{2}}\).