Question

The total cost function of a firm is C(x) = ${{x}^{2}}+75x+1600$ for output x. find the output ( x ) for which average cost is minimum. Is ${{C}_{A}}={{C}_{M}}$ at this output?

Answer 1

Hint: We have to find the output ( x ) for which average cost function is minimum and we have to check whether ${{C}_{A}}={{C}_{M}}$ for output we obtained where ${{C}_{A}}$ denotes Average Cost Function and ${{C}_{M}}$ denotes Marginal cost function. We will use formulas of Average cost function denoted by${{C}_{A}}$ and equals to ${{C}_{A}}(x)=\dfrac{C(x)}{x}$ and Marginal cost function denoted by ${{C}_{M}}$ and equals to ${{C}_{m}}=C'(x)$.

Complete step by step solution:
Now, first we’ll calculate the value of Average cost function which is given by the formula ${{C}_{A}}(x)=\dfrac{C(x)}{x}$…… ( i )
In question we have, C(x) = ${{x}^{2}}+75x+1600$…… (ii)
 Substituting value of ( ii ) in ( i ), we get
${{C}_{A}}(x)=\dfrac{{{x}^{2}}+75x+1600}{x}=x+75+\dfrac{1600}{x}$
Now, minimum average cost is obtained by ${{C}_{A}}'(x)=0$and also ${{\bar{C}}_{A}}''(x)>0$ . Differentiating $\bar{C}(x)$ with respect to x, we get ${{C}_{A}}'(x)=1-\dfrac{1600}{{{x}^{2}}}$……. ( iii )
As $\dfrac{d}{dx}(k)=0,\dfrac{d}{dx}({{x}^{n}})=n{{x}^{n-1}},\dfrac{d}{dx}(\dfrac{1}{{{x}^{n}}})=-n\cdot \dfrac{1}{{{x}^{n+1}}}$ ; where k is any constant and n $\in $ Real number
Putting ( iii ) equals to 0, we get
$\begin{align}
  & 1-\dfrac{1600}{{{x}^{2}}}=0 \\
 & {{x}^{2}}=1600 \\
 & x=40 \\
\end{align}$
And, ${{C}_{A}}''(x)=\dfrac{3200}{{{x}^{3}}}>0$ for x = 40.
So, at x = 40 we get minimum average cost.
Minimum Average Cost = ${{C}_{A}}(x)=40+75+\dfrac{1600}{40}=155$
$\therefore $ ${{C}_{A}}=155$
Marginal cost function = ${{C}_{m}}=C'(x)$
${{C}_{m}}=\dfrac{d}{dx}({{x}^{2}}+75x+1600)$
 ${{C}_{m}}=2x+75$…… ( iv )
Putting x = 40 in ( iv )
$\therefore $ ${{C}_{m}}=155$
We can see that, ${{C}_{A}}={{C}_{M}}$ for x = 40
Hence, minimum average cost value is equals to 155 at x = 40 and also we get marginal cost value equals to 155 at x = 40 and we can see ${{C}_{A}}={{C}_{M}}$ for x = 40.

Note: This is the direct way to find the values of average cost function and marginal cost function at any output ( x ). Differentiate functions carefully with respect to x. be careful while handling formulas of cost and differentiation as it may give you incorrect numerical value.