Hello Marty,
I would first observe that since $x$ represents a distance, we will require:
$$0\le x$$
Next, let's look at a graph of the function where $x\le x\le50$:
View attachment 1439
As we can see, the cost function is minimized when:
$$x\approx20$$
Now, let's find this critical value for $x$.
i) First we will use a pre-calculus method:
$$C=\frac{x^2+400}{10x}$$
$$x^2-10Cx+400=0$$
The axis of symmetry, where the vertex of the quadratic is located, is given by:
$$x=-\frac{-10C}{2(1)}=5C$$
Substituting this into the quadratic, we find:
$$(5C)^2-10C(5C)+400=0$$
(5C)^2=20^2
Taking the positive root, we find:
$$5C=20$$
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:
$$ C(x)=\frac{40}{x}+\frac{x}{10}=40x^{-1}+\frac{1}{10}x$$
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).
$$C'(x)=-40x^{-2}+\frac{1}{10}=\frac{x^2-400}{10x^2}=0$$
We know the cost function grows unbounded as $x$ approaches zero, and so we are interesting only in the critical values from the numerator:
$$x^2-400=0$$
$$x^2=20^2$$
Taking the positive root, we obtain:
$$x=20$$
Using the second derivative test, we find:
$$C''(x)=80x^{-3}$$
Now since $$C''(20)>0$$ we can conclude the cost function is concave up at this critical value, and so we know we have found the global minimum.
