The total revenue and the total cost (in Rs.) of producing and selling x units of an item are \(R(x) = 60x - x^{2}\) and \(C(x) = x^{2} + 12x + 50\). Match List-I with List-II.
| List-I | List-II | ||
| (A) | Marginal cost when x = 4 (in Rs.) | (I) | 20 |
| (B) | Marginal revenue when x = 10 (in Rs.) | (II) | 12 |
| (C) | Number of units for maximum profit | (III) | 40 |
| (D) | Maximum profit (in Rs.) | (IV) | 238 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(I), (B)-(III), (C)-(II), (D)-(IV)
The marginal cost and marginal revenue are the derivatives C′(x) and R′(x), and the profit is P(x) = R(x) − C(x):
\( C'(x) = 2x + 12,\ R'(x) = 60 - 2x \)
\( P(x) = -2x^{2} + 48x - 50 \)
(A) Marginal cost when x = 4 (in Rs.)
This matches (I) 20.
\( C'(4) = 8 + 12 = 20 \)
(B) Marginal revenue when x = 10 (in Rs.)
This matches (III) 40.
\( R'(10) = 60 - 20 = 40 \)
(C) Number of units for maximum profit
This matches (II) 12.
\( P'(x) = -4x + 48 = 0,\ x = 12;\ P'' = -4 \lt 0 \)
(D) Maximum profit (in Rs.)
This matches (IV) 238.
\( P(12) = -288 + 576 - 50 = 238 \)