An oil refinery is located 1 km north of the north bank of a straight river that is 3 km wide. A pipeline is to be constructed from the refinery to storage tanks located on the south bank of the river 5 km east of the refinery. The cost of laying pipe is $200,000/km over land to a point P on the north bank and $400,000/km under the river to the tanks. To minimize the cost of the pipeline, how far downriver from the refinery should the point P be located?

Answer:The point should be placed 3.362 km to the east of the refinery to minimize the cost.Step-by-step explanation:You should draw a graph that depicts the situation.Let x be the distance from the refinery to the point P on the north bank in km and C the total cost of the pipeline.According to the graph and using Pythagoras's theorem, the distance that the pipe should cover is given by[tex]D=\sqrt{x^2+1^2}+\sqrt{(5-x)^2+3^2}  \\D=\sqrt{x^2+1}+\sqrt{25-10x+x^2+9} \\D=\sqrt{x^2+1}+\sqrt{x^2-10x+34}[/tex]We know that the cost of laying pipe is $200,000/km over land to a point P on the north bank and $400,000/km under the river to the tanks. Therefore, the total cost of the pipeline is the cost of the part on land plus the cost of the amount underwater.[tex]C(x)=200000\sqrt{x^2+1}+400000\sqrt{x^2-10x+34}[/tex]The domain of C(x) is [0, 5].To find the minimum cost, we need to differentiate the cost function, set it to zero and solve for x to find the critical points.[tex]\frac{d}{dx} C(x)=\frac{d}{dx} (200000\sqrt{x^2+1}+400000\sqrt{x^2-10x+34})\\\\C'(x)=\frac{d}{dx}\left(200000\sqrt{x^2+1}\right)+\frac{d}{dx}\left(400000\sqrt{x^2-10x+34}\right)\\\\C'(x)=\frac{200000x}{\sqrt{x^2+1}}+\frac{200000\left(2x-10\right)}{\sqrt{x^2-10x+34}}[/tex][tex]\frac{200000x}{\sqrt{x^2+1}}+\frac{200000\left(2x-10\right)}{\sqrt{x^2-10x+34}}=0[/tex]Solving this equation with the help of graphing calculator (Desmos) we get that[tex]x=3.362[/tex]We can use The Second Derivative Test to check if x is a minimum,if C'(c) = 0 and C''(c) > 0, then C has a local minimum at c.The second derivative is[tex]\frac{d}{dx} C'(x)=\frac{d}{dx}(\frac{200000x}{\sqrt{x^2+1}}+\frac{200000\left(2x-10\right)}{\sqrt{x^2-10x+34}})\\\\C''(x)=\frac{d}{dx}\left(\frac{200000x}{\sqrt{x^2+1}}\right)+\frac{d}{dx}\left(\frac{200000\left(2x-10\right)}{\sqrt{x^2-10x+34}}\right)\\\\C''(x)=\frac{200000}{\left(x^2+1\right)\sqrt{x^2+1}}+\frac{3600000}{\left(x^2-10x+34\right)\sqrt{x^2-10x+34}}[/tex]Next, substitute [tex]x=3.362[/tex] into the second derivative[tex]\frac{200000}{\left(3.362^2+1\right)\sqrt{3.362^2+1}}+\frac{3600000}{\left(3.362^2-10\left(3.362\right)+34\right)\sqrt{3.362^2-10\left(3.362\right)+34}}=94785[/tex]Because C''(3.362) > 0, then C(x) has a local minimum at 3.362.Thus the point should be placed 3.362 km to the east of the refinery to minimize the cost.