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Q.If the cost function and revenue function of xx items are respectively given as C(x)=100+0.015 x2C(x) = 100 + 0.015\,x^2, R(x)=3xR(x) = 3x, then the value of xx for maximum profit is (A) 5050 (B) 100100 (C) 150150 (D) 200200

CBSECBSE Class XII Board 2025MCQ· 1mImportance★★★★★
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P(x)=3x−100−0.015x2P(x)=3x-100-0.015x^2; P′(x)=3−0.03x=0⇒x=100P'(x)=3-0.03x=0\Rightarrow x=100, a maximum since P′′(x)=−0.03<0P''(x)=-0.03<0.

Profit P(x)=R(x)−C(x)P(x)=R(x)-C(x); maximum where P′(x)=0P'(x)=0 and P′′(x)<0P''(x)<0.

  1. Form the profit function: P(x)=R(x)−C(x)=3x−(100+0.015x2)=3x−100−0.015x2P(x)=R(x)-C(x)=3x-(100+0.015x^2)=3x-100-0.015x^2.
  2. Differentiate: P′(x)=3−0.03xP'(x)=3-0.03x. …

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