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Class 12 Applications Of Derivatives · Very Long Answer5 Marksprofit_optimization

Class 12 Maths · Applications Of Derivatives · Very Long Answer #106

A manufacturer sells x items at ₹(5−x/100) each. Their total cost is ₹(x/5+500). Find the number of items maximizing profit, for integer 0≤x≤500.

✅ Step-by-Step Solution

Final Answer: 240 items; maximum profit ₹76.
  1. Revenue R=x(5−x/100); profit P=R−(x/5+500).
  2. Thus P(x)=24x/5−x²/100−500.
  3. P′=24/5−x/50=0 gives x=240.
  4. P″=−1/50<0, so the vertex is the global maximum and 240 is feasible and integral.
  5. Sell 240 items; P(240)=1152−576−500=₹76.

📋 CBSE Stepwise Marking Scheme

  • [1 mark] Revenue R=x(5−x/100); profit P=R−(x/5+500).
  • [1 mark] Thus P(x)=24x/5−x²/100−500.
  • [1 mark] P′=24/5−x/50=0 gives x=240.
  • [1 mark] P″=−1/50<0, so the vertex is the global maximum and 240 is feasible and integral.
  • [1 mark] Sell 240 items; P(240)=1152−576−500=₹76.

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