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3.5 · Q7

Q.If price 'p' per unit of an article is p = 75 – 2x and the cost function is C(x)=350+12x+x24C(x) = 350 + 12x + \dfrac{x^2}{4}. Find the number of units and the price at which the total profit is maximum. What is the maximum profit?

Puducherry CbseNCERTSubjective· 3mImportance★★★★★
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Build profit P=R−C=63x−94x2−350P=R-C=63x-\tfrac94 x^2-350; maximising gives x=14x=14 units, price ₹47, and maximum profit ₹91.

Revenue R=p⋅xR=p\cdot x; Profit P(x)=R(x)−C(x)P(x)=R(x)-C(x); maximum profit at P′(x)=0P'(x)=0 with P′′(x)<0P''(x)<0.

  1. Given p=75−2xp=75-2x and C(x)=350+12x+x24C(x)=350+12x+\dfrac{x^2}{4}.
  2. Revenue: R(x)=p⋅x=(75−2x)x=75x−2x2R(x)=p\cdot x=(75-2x)x=75x-2x^2.
  3. Profit: P(x)=R−C=(75x−2x2)−(350+12x+x24)=63x−94x2−350P(x)=R-C=(75x-2x^2)-\left(350+12x+\dfrac{x^2}{4}\right)=63x-\dfrac{9}{4}x^2-350 (since 2x2+14x2=94x22x^2+\tfrac14x^2=\tfrac94x^2).
  4. Differentiate: P′(x)=63−92xP'(x)=63-\dfrac{9}{2}x.
  5. Set P′(x)=0⇒92x=63⇒x=1269=14P'(x)=0\Rightarrow \dfrac{9}{2}x=63\Rightarrow x=\dfrac{126}{9}=14.
  6. Second derivative: P′′(x)=−92<0P''(x)=-\dfrac{9}{2}<0, so x=14x=14 gives maximum profit.
  7. Price: p=75−2(14)=75−28=₹47p=75-2(14)=75-28=₹47. …

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