<h2>
Answer:</h2>
<em><u>Distance of the pipeline down the shoreline should be = 3.34 miles</u></em>
<h2>
Step-by-step explanation:</h2>
In the question,
Distance of the Resort, R, from the point, P = 5 miles
Distance of the Fresh Source of water, X, from P = 10 miles
Also, if,
Cost of laying pipeline on land = 1
then,
Cost of laying pipeline in water = 1.8
So,
Using Pythagoras theorem in triangle PRX, we get,
Length of pipe in water, LR is,
![LR^{2}=PL^{2}+PR^{2}\\LR^{2}=x^{2}+25\\LR=\sqrt{x^{2}+25}](https://tex.z-dn.net/?f=LR%5E%7B2%7D%3DPL%5E%7B2%7D%2BPR%5E%7B2%7D%5C%5CLR%5E%7B2%7D%3Dx%5E%7B2%7D%2B25%5C%5CLR%3D%5Csqrt%7Bx%5E%7B2%7D%2B25%7D)
So,
<u>Total cost, C, of laying the pipeline</u> is,
![C=1.(10-x)+1.8(\sqrt{x^{2}+25})](https://tex.z-dn.net/?f=C%3D1.%2810-x%29%2B1.8%28%5Csqrt%7Bx%5E%7B2%7D%2B25%7D%29)
On differentiating it w.r.t x, we get,
![\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](https://tex.z-dn.net/?f=%5Cfrac%7BdC%7D%7Bdx%7D%3D-1%2B%5Cfrac%7B1.8x%7D%7B%5Csqrt%7Bx%5E%7B2%7D%2B25%7D%7D%5C%5C0%20%3D%20-1%2B%5Cfrac%7B1.8x%7D%7B%5Csqrt%7Bx%5E%7B2%7D%2B25%7D%7D%5C%5C1%3D%5Cfrac%7B1.8x%7D%7B%5Csqrt%7Bx%5E%7B2%7D%2B25%7D%7D%5C%5Cx%5E%7B2%7D%2B25%3D3.24x%5E%7B2%7D%5C%5C2.24x%5E%7B2%7D%3D25%5C%5Cx%5E%7B2%7D%3D11.16%5C%5Cx%3D3.34%5C%2Cmiles)
<em><u>Therefore, the distance of the pipeline down the shoreline should be,</u></em>
<em><u>x = 3.34 miles to minimize the construction cost.</u></em>