# 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.