Single Variable Calculus, Chapter 4, 4.7, Section 4.7, Problem 48

Suppose that the Oil Refinery is located 1k north of the river that is 2km wide. A pipeline is to be contructed from the refinery to storage tanks located on the south bank of the river 6km east of the refinery. The cost of laying pipe is $\displaystyle \frac{\$ 400,000}{\text{km}}$ over land to a point $P$ on the north bank and $\displaystyle \frac{\$800,000}{\text{km}}$ under the river to the tanks. Where should $P$ be located to minimize the cost of the pipeline?



By Pythagorean Theorem, the distance between the refinery to point $P$ is ...
$d_1 = \sqrt{(6-x)^2+1^2} = \sqrt{x^2-12x+37}$

And, the distance between point $P$ to the storage is...
$d_2 = \sqrt{x^2 +4}$

Therefore, the total cost would be...
cost = $ 400,000 \sqrt{x^2 - 12x + 37} + 800,000\sqrt{x^2+4}$

If we take the derivative of cost, we have...
$\displaystyle c'(x) = 400,000 \left( \frac{2x-12}{2\sqrt{x^2-12x+37}} \right) + 800,000 \left( \frac{2x}{2\sqrt{x^2+4}} \right)$

when $c'(x) = 0$

$
\begin{equation}
\begin{aligned}
0 &= 400,000 \left[ \frac{\cancel{2}(x-6) }{\cancel{2}\sqrt{x^2 - 12x + 37} } + 2 \left( \frac{\cancel{2}x }{\cancel{2}\sqrt{x^2 + 4} } \right) \right]\\
\\
0 &= \frac{x - 6}{\sqrt{x^2 - 12x + 37}} + \frac{2x}{\sqrt{x^2 + 4}}
\end{aligned}
\end{equation}
$


Solving for $x$, we have...
$x = 1.12$ km

If we evaluate the cost at $x = 0$, $x = 6$km and $x = 1.12$km, then...

$
\begin{equation}
\begin{aligned}
c (0) &= 400,000 \left( \sqrt{0^2 - 12(0) +37} \right) + 800,000 \left( \sqrt{0^2 + 4} \right)\\
\\
c (0) &= \$ 4,033,105.012\\
\\
c (6) &= 400,00 \left( \sqrt{6^2 - 12(6)+37} \right) + 800,000 \left( \sqrt{6^2+4} \right)\\
\\
c (0) &= \$ 5,459,644.256\\
\\
c (1.12) &= 400,000 \left( \sqrt{(1.12)^2 - 12(1.12) + 37 } \right) + 800,000 \left( \sqrt{(1.12)^2 + 4} \right)\\
\\
c (1.12) &= \$ 3,826,360.414

\end{aligned}
\end{equation}
$