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Minimizing pipeline cost

A utility company must run a pipeline from a station on one side of a straight river to a plant on the other side. The river is 44 miles wide, and the plant is located 2020 miles downstream from the point directly across the river from the station. The pipeline will run underwater in a straight line from the station to some point on the opposite bank, and then along the bank (over land) to the plant. Let xx be the distance (in miles), measured along the bank, from the point directly across from the station to the point where the pipeline comes ashore. Underwater pipe costs \6millionpermile,andlandpipecostsmillion per mile, and land pipe costs$4$ million per mile.

The total cost of the pipeline (in millions of dollars) as a function of xx is C(x)=6x2+16+4(20−x).C(x) = 6\sqrt{x^2+16} + 4(20-x). Here xx is the distance along the bank from the point directly opposite the station to where the pipe reaches shore, and 0≤x≤200 \le x \le 20. Find the value of xx that minimizes the total cost C(x)C(x).

Hint

Take the derivative C′(x)C'(x), set it equal to zero, and solve for xx. Since x2+16>0\sqrt{x^2+16} > 0 always, be careful when clearing the square root — check that your solution satisfies the original equation (i.e., has the correct sign) and lies in the domain [0,20][0,20].

Step-by-step solution

Differentiate C(x)=6x2+16+4(20−x)C(x) = 6\sqrt{x^2+16} + 4(20-x) to get C′(x)=6xx2+16−4C'(x) = \dfrac{6x}{\sqrt{x^2+16}} - 4. Setting C′(x)=0C'(x)=0 gives 6xx2+16=4\dfrac{6x}{\sqrt{x^2+16}} = 4, so 6x=4x2+166x = 4\sqrt{x^2+16}, requiring x>0x>0.

Squaring both sides: 36x2=16(x2+16)36x^2 = 16(x^2+16), so 20x2=25620x^2 = 256, giving x2=12.8x^2 = 12.8 and x=12.8≈3.578x = \sqrt{12.8} \approx 3.578. Since C′(x)<0C'(x)<0 for x<12.8x<\sqrt{12.8} and C′(x)>0C'(x)>0 for x>12.8x>\sqrt{12.8} (or by checking C′′(x)>0C''(x)>0), this critical point is a minimum, and it lies in [0,20][0,20], so it is the answer.

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