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Marty's question at Yahoo! Answers regarding minimizing a cost function

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MarkFL

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Feb 24, 2012
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Here is the question:

Help with cost functions?


Hi there,

Got this cost function question where C(x) = 40/x + x/10
I need to find x where x=distance in meters between the poles that will minimize the cost.

Thanks for any help/hints!
I have posted a link there to this topic so the OP can see my work.

edit: Unfortunately, the OP deleted the question before I had a chance to post my response there.
 
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MarkFL

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Feb 24, 2012
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Hello Marty,

I would first observe that since $x$ represents a distance, we will require:

\(\displaystyle 0\le x\)

Next, let's look at a graph of the function where $x\le x\le50$:

marty.jpg

As we can see, the cost function is minimized when:

\(\displaystyle x\approx20\)

Now, let's find this critical value for $x$.

i) First we will use a pre-calculus method:

\(\displaystyle C=\frac{x^2+400}{10x}\)

\(\displaystyle x^2-10Cx+400=0\)

The axis of symmetry, where the vertex of the quadratic is located, is given by:

\(\displaystyle x=-\frac{-10C}{2(1)}=5C\)

Substituting this into the quadratic, we find:

\(\displaystyle (5C)^2-10C(5C)+400=0\)

(5C)^2=20^2

Taking the positive root, we find:

\(\displaystyle 5C=20\)

And this is our critical value. Since the cost function grows unbounded as $x$ approaches zero and as $x$ approaches infinity, we may conclude this is a global minimum on the applicable domain.

ii) Next, let's apply the calculus:

\(\displaystyle C(x)=\frac{40}{x}+\frac{x}{10}=40x^{-1}+\frac{1}{10}x\)

Now, in order to find the extrema, we need to compute the first derivative, and equate it to zero, and solve for $x$ to get the critical value(s).

\(\displaystyle C'(x)=-40x^{-2}+\frac{1}{10}=\frac{x^2-400}{10x^2}=0\)

We know the cost function grows unbounded as $x$ approaches zero, and so we are interesting only in the critical values from the numerator:

\(\displaystyle x^2-400=0\)

\(\displaystyle x^2=20^2\)

Taking the positive root, we obtain:

\(\displaystyle x=20\)

Using the second derivative test, we find:

\(\displaystyle C''(x)=80x^{-3}\)

Now since \(\displaystyle C''(20)>0\) we can conclude the cost function is concave up at this critical value, and so we know we have found the global minimum.
 

Jameson

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Jan 26, 2012
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edit: Unfortunately, the OP deleted the question before I had a chance to post my response there.
Bummer :( Thank you for posting this anyway and for all of the questions you bring in!
 
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MarkFL

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Feb 24, 2012
13,775
Bummer :( Thank you for posting this anyway and for all of the questions you bring in!
Yeah, I was mildly annoyed at first, but I just decided to do another one. (Rofl)