Cost optimization problem. Marginal cost is just the derivative of the cost function right

Question

I have this question:

a) If $C(x)$ is the cost of producing $x$ units of a commodity, then the average cost per unit is $c(x) = \dfrac{C(x)}{x}$. Show that if the average cost is a minimum, then the marginal cost equals the average cost.

So for a):

Marginal Cost: $C'(x)$, right?

Avg cost = $\frac{C(x)}{x}$

And so avg cost':

$$ \frac{x \cdot C'(x) - C(x)}{x^2}$$

$$ \frac{C'(x)}{x} - \frac{C(x)}{x^2}$$

But then the marginal cost does not equal the derivative of the avg cost when minimized.

Basically, I'm trying to set the derivative of the avg cost function to 0 to find some critical values, values at which a minimum could occur and see if it equals marginal cost. But I'm a bit stuck...

Answer 1

a) You are a step away from answer: $$\frac{xC'(x)-C(x)}{x^2}\color{red}{=0} \Rightarrow xC'(x)-C(x)=0 \Rightarrow C'(x)=\frac{C(x)}{x}.$$

Answer 2

Note that the average cost is minimized when its derivative is $0$

$$ \big (\frac{C(x)}{x}\big)'=\frac{x \cdot C'(x) - C(x)}{x^2}=0 \implies $$

$$C'(x)= \frac{C(x)}{x}$$

That is the marginal cost equals the average cost.

The rest of the problem is not an issue.

Answer 3

Apply Quotient Rule to average cost when differentiating to find out when the cost reached a minimum.

After simplifying the quotient

$$ \dfrac{C(x)}{x} = \dfrac{C^{'}(x)}{1} $$

which is nothing but marginal cost by its very definition.