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CA Foundation · Quantitative Aptitude · Differential and Integral Calculus

The profit of a Pune firm is P(x) = −x³ + 12x² + 60x − 200 (in ₹ thousand) for output x units, x > 0. The output at which profit is maximum is:

Profit is maximum at x = 10 units. Setting P'(x) = −3x² + 24x + 60 equal to zero gives x² − 8x − 20 = 0, so x = 10 for positive output, and P''(10) = −36 is negative, confirming a maximum.

  1. Ax = 10Correct
  2. Bx = 6
  3. Cx = 12
  4. Dx = 5

Explanation

P'(x) = −3x² + 24x + 60 = 0 gives x² − 8x − 20 = 0, so (x − 10)(x + 2) = 0 and x = 10 (as x > 0). P''(x) = −6x + 24 = −36 < 0 at x = 10, so it is a maximum. x = 6 is the point where P'' = −12... actually P''(x)=0 at x = 4, not 6, so 6 is not a stationary point.

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