In this post we are going to look at a function and find its minimum value analytically. Instead of just a dry exercise, it is usually more fun to tackle a word problem. All calculus textbooks have several versions of this very same problem.
The problem: An island is 3.5 miles from the nearest point on a straight shoreline; that point is 8 miles from a processing station (see figure). An energy company plans to lay a gas pipeline underwater from the island to the shore and then underground along the shore to the processing station. Assume it costs $2400 per mile to lay an underwater pipeline and $1200 per mile to lay an underground pipeline. At what point should the underwater pipeline meet the shore in order to minimize the cost of the project? One option is to build the pipeline with the underwater portion as short as possible: from the island to point a on the shore and the rest on land. Another option is to build it entirely underwater, straight from the island to the processing station. Finally, it could be built from the island to the shore somewhere between point a and the station.
I have found that in this kind of problems, it is helpful to look for geometric shapes. The right triangle with sides island<–>a, a<–>b and b<–>island (hypothenuse) is pretty obvious.
Perhaps the best way to proceed is to divide the shoreline in two parts:
\(\boldsymbol{x} = \) the distance from the point on the shore closest to the island, to the location where the underwater pipeline reaches the shore (a -> b on the figure).
\(\boldsymbol{8\, – x} = \) the remaining shoreline; that is, from where the underwater pipeline reaches the shore to the processing station (b -> c on the figure).
Once we do that, we can build some equations:
\begin{alignat}{2}
\text{pWater} &= \sqrt{\left(3.5\right)^{2} + x^{2}} \qquad & \text{(underwater pipeline)} \\
\text{pLand} &= 8\, – x & \text{(pipeline on land)} \\
\text{pCost} &= 2400 \sqrt{\left(3.5\right)^{2} + x^{2}} + 1200 \left(8\, – x\right) & \text{(cost function)} \\
&= 2400 \sqrt{\frac{49}{4} + x^{2}}\, – 1200x + 9600
\end{alignat}
Now we optimize the cost function (\(\boldsymbol{pCost}\)); that is, we compute the function minimum. How do we do that? Well, that is one of the uses of Differential Calculus!
\begin{align}
\text{Take the derivative of the cost function:} \frac{d}{d x} \left(2400 \sqrt{\frac{49}{4} + x^{2}}\, – 1200x +9600\right) &= \frac{4800x}{\sqrt{4x^{2} + 49}}\, – 1200 \\
\text{Set it equal to zero and solve for } x \text{:}\quad \frac{4800x}{\sqrt{4x^{2} + 49}}\, – 1200 &= 0 \\
x &= \frac{7\sqrt{3}}{6}\, \approx 2.0208
\end{align}
\(\boldsymbol{x = \frac{7\sqrt{3}}{6}}\) is a critical point; the location where the slope of the tangent line is zero. But is this a minimum or a maximum? The easiest way to proceed is to check the value of the cost function at \(\boldsymbol{x = \frac{7\sqrt{3}}{6}}\) and also at the function endpoints; that is, when \(\boldsymbol{x = 0 \text{ or } x = 8}\).
\begin{align}
\text{Underwater pipeline to point (a) and then 8 miles on land to plant: } pCost\left(0\right) &= $18,000.00 \\
\text{Underwater pipeline to point (b) and then on land to plant: } pCost\left(\frac{7\sqrt{3}}{6}\right) &= $16,874.61 \\
\text{Pipeline entirely underwater, from the island all the way to plant: } pCost\left(8\right) &= $20,957.10
\end{align}
A look at the graph of the Cost Function:
If you find any errors in this post, or have some thoughts about it, please, email me at [email protected] .