A small resort is situated on an island that lies exactly 5 miles from P, the nearest point to the island along a perfectly straight shoreline. 10 miles down the shoreline from P is the closest source of fresh water. If it costs 1.8 times as much money to lay pipe in the water as it does on land, how far down the shoreline from P should the pipe from the island reach land in order to minimize the total construction costs?

Answer:Distance of the pipeline down the shoreline should be = 3.34 milesStep-by-step explanation:In the question,Distance of the Resort, R, from the point, P = 5 milesDistance of the Fresh Source of water, X, from P = 10 milesAlso, if,Cost of laying pipeline on land = 1then,Cost of laying pipeline in water = 1.8So,Using Pythagoras theorem in triangle PRX, we get,Length of pipe in water, LR is,[tex]LR^{2}=PL^{2}+PR^{2}\\LR^{2}=x^{2}+25\\LR=\sqrt{x^{2}+25}[/tex]So,Total cost, C, of laying the pipeline is, [tex]C=1.(10-x)+1.8(\sqrt{x^{2}+25})[/tex]On differentiating it w.r.t x, we get,[tex]\frac{dC}{dx}=-1+\frac{1.8x}{\sqrt{x^{2}+25}}\\0 = -1+\frac{1.8x}{\sqrt{x^{2}+25}}\\1=\frac{1.8x}{\sqrt{x^{2}+25}}\\x^{2}+25=3.24x^{2}\\2.24x^{2}=25\\x^{2}=11.16\\x=3.34\,miles[/tex]Therefore, the distance of the pipeline down the shoreline should be,x = 3.34 miles to minimize the construction cost.