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

#### MarkFL

Staff member
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.

#### MarkFL

Staff member
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 $x\le x\le50$: 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

Bummer Thank you for posting this anyway and for all of the questions you bring in!
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. 