Let C = C(x) be the total cost of producing x units of a product and AC is the average cost and MC is the marginal cost, then
Match List-I with List-II
| List-I | List-II | ||
| (A) | AC decreases with x | (I) | when MC = AC |
| (B) | AC increases with x | (II) | when MC < AC |
| (C) | AC doesn't change with x | (III) | when MC = 0 |
| (D) | C is constant. | (IV) | when AC < MC |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (II), (B) - (IV), (C) - (I), (D) - (III)
Here \( AC=\frac{C}{x} \) and \( MC=\frac{dC}{dx} \). By the quotient rule:
\( \displaystyle \frac{d}{dx}(AC)=\frac{x\frac{dC}{dx}-C}{x^2} \)
\( \displaystyle =\frac{1}{x}\left(\frac{dC}{dx}-\frac{C}{x}\right) \)
\( \displaystyle =\frac{MC-AC}{x} \)
Since x > 0, the sign of the change in AC is the sign of MC − AC.
(A) AC decreases with x
This matches (II) when MC < AC.
MC − AC < 0 makes \( \frac{d}{dx}(AC)\lt 0 \).
(B) AC increases with x
This matches (IV) when AC < MC.
MC − AC > 0 makes \( \frac{d}{dx}(AC)\gt 0 \).
(C) AC doesn't change with x
This matches (I) when MC = AC.
MC − AC = 0 makes \( \frac{d}{dx}(AC)=0 \).
(D) C is constant.
This matches (III) when MC = 0.
A constant total cost has zero derivative: \( MC=\frac{dC}{dx}=0 \).
So the matching is (A)-(II), (B)-(IV), (C)-(I), (D)-(III).