MHT CET · Maths · Application of Derivatives
A manufacturer sells \(x\) items at a price of rupees \(\left(6-\frac{x}{40}\right)\) each. The cost price of \(x\) items is Rs. \(\left(\frac{x}{5}+193\right)\). The maximum profit in Rs. _______ is
- A 134.4
- B 144.3
- C 143.4
- D 133.4
Answer & Solution
Correct Answer
(C) 143.4
Step-by-step Solution
Detailed explanation
Revenue \(R(x) = x\left(6-\frac{x}{40}\right) = 6x - \frac{x^2}{40}\) Profit \(P(x) = R(x) - C(x) = \left(6x - \frac{x^2}{40}\right) - \left(\frac{x}{5}+193\right)\)
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