Oil Rig Optimization

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jetter2

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Nov 2, 2010
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I set this problem up and got a value of X to be 3.25. We are optimizing the cost, and solving for how far point B is from the Refinery

I was wrong and the x value should be x= 12/sqrt(7)

..can someone else crank this one out and let me see how they did it?

calc.JPG
 
jetter2 said:
I set this problem up and got a value of X to be 3.25. We are optimizing the cost, and solving for how far point B is from the Refinery

I was wrong and the x value should be x= 12/sqrt(7)

..can someone else crank this one out and let me see how they did it?

calc.JPG
Click to expand...

You did not tell us what did you deine as "x". Please tell us how did you get x = 3.25.

Please share your work with us - so that we may know where to begin to help you.
 
X = Distance from Point B to Oil Refinary. Its a Cost optimization problem.

1)Create a cost function
2)Take its derivative
3)Set to Zero
4)Solve for X

I will work out what I can and post shortly.
 
jetter2 said:
X = Distance from Point B to Oil Refinary. Its a Cost optimization problem.

1)Create a cost function
2)Take its derivative
3)Set to Zero
4)Solve for X

I will work out what I can and post shortly.
Click to expand...

Actually for the answer to be x = 12/?7, the (9-x) is the distance between B and the refinery.
 
If 9-x were on the triangle, you would have a nastier equation. But yes, you could do it either way.

You will get a different solution the other way. But it would eventually work out to the same total cost.

The reason is because x is a different length in each case.

Case 1: x is the base of the triangle and 9-x is the pipe on land:

\(\displaystyle C(x)=300(9-x)+400\sqrt{x^{2}+16}\)

\(\displaystyle C'(x)=\frac{400x-300\sqrt{x^{2}+16}}{\sqrt{x^{2}+16}}\)

Set \(\displaystyle 400x-300\sqrt{x^{2}+16}=0\).............solve for x.

Now, the other way.

Case #2: 9-x is the base of the triangle and x is the pipe on land.

\(\displaystyle C(x)=300x+400\sqrt{x^{2}-18x+97}\)

\(\displaystyle C'(x)=\frac{300\sqrt{x^{2}-18x+97}+400x-3600}{\sqrt{x^{2}-18x+97}}\)

\(\displaystyle 300\sqrt{x^{2}-18x+97}+400x-3600=0\)..................solve for x.

Which would you rather solve?.

Either way, sub your solution back into the cost function to find the total cost.
 
Like Galactus explained to you - you should get same "minimum cost" either way.

If you have to differentiate - and if you have "radical" sign - and if you have a choice - choose the simplest expression under the radicals.
 
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