MHB Building a Pipeline Optimization Problem

[ardentmed]

Hey guys, I'm having trouble with this problem set I'm working on at the moment. I'd appreciate some help with this question:

(I'm only asking about number two. Ignore number one please.)

So if the length for the hypotenuse of the leftmost triangle is represented by:
c^2 = x^2 + y^2

Then,

c= √(2500 + x^2)

Therefore, the total cost comes to:

C(x) = 400,000-20,000x + 50,000√(2500 + x^2)

Am I on the right track?

Moreover, we need to optimize and deduce the minimum cost, x's smallest possible value:

c'(x) = dy/dx (400,000-20,000x + 50,000√(2500 + x^2))

Then isolate and solve for "x."

x=21.821789 km

x ~ 21.km.

Thanks in advance.

 

[MarkFL]

I like to work problems like this in general terms, so that I have a formula to use for future problems. Let's let the distance of the oil rig from the shore be $O$, the distance down shore from the rig to the plant be $P$, the cost to per unit length to lay the pipe over land be $C$ and the cost per unit length to lay the pipe underwater be $kC$ where $0\le k,\,k\ne 1$. We will then let $x$ be the point on the shoreline we select to minimize the cost of the pipeline.

Hence, the total cost function $C_T$ is:

$$C_T(x)=C(P-x)+kC\sqrt{x^2+O^2}=C\left(P-x+k\sqrt{x^2+O^2}\right)$$

We need only optimize the factor containing $x$. THus differentiating, and equating the result to zero, we find:

$$\frac{d}{dx}\left(\frac{C_T}{C}\right)=\frac{kx}{\sqrt{x^2+O^2}}-1=0$$

This implies:

$$x=\frac{O}{\sqrt{k^2-1}}$$

Now, observing that:

$$\frac{d^2}{dx^2}\left(\frac{C_T}{C}\right)=\frac{kO^2}{\left(x^2+O^2\right)^{\frac{3}{2}}}>0$$

for all real $x$, we know our critical point is at a minimum. So, we can now plug in the given data (distances in km):

$$O=50,\,k=\frac{5}{2}$$

we obtain:

$$x=\frac{50}{\sqrt{\left(\frac{5}{2}\right)^2-1}}=\frac{100}{\sqrt{21}}\approx21.82$$

So, our answers agree. (Yes)