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MHT CET202526 Apr 2025Morning ShiftMathematicsApplication of DerivativesActual

A manufacturer sells x items at a price of rupees (6- x 40 ) each. The cost price of x items is Rs. ( x 5 +193 ) . The maximum profit in Rs. _______ is

Options

  1. A134.4
  2. B144.3
  3. C143.4
  4. D133.4

Correct answer

C. 143.4

Step-by-step solution

Let x be the quantity sold. The revenue function is R(x) = x P(x) = x (6 - x 40 ) = 6x - x^2 40 . The cost is given by C(x) = x 5 + 193 , so the profit function becomes: Profit (x) = R(x) - C(x) = 6x - x^2 40 - x 5 - 193 Combining like terms yields: Profit (x) = - x^2 40 + 29 5 x - 193 The derivative with respect to x is: d dx Profit (x) = - x 20 + 29 5 Setting the derivative to zero gives: - x 20 + 29 5 = 0 Solving, x = 116 is the critical point. Since the second derivative is d^2 dx^2 Profit (x) = - 1 20 Substitu

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