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3.5 · Q6

Q.A tour operator charges ₹136 per passenger for 100 passengers with a discount of ₹4 for each 10 passengers in excess of 100. Find the number of passengers that will maximise the amount of money the tour operator receives.

Chandigarh CbseNCERTSubjective· 3mImportance★★★★★
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Model the fare drop in blocks of 10 excess passengers; maximising revenue R=(100+10x)(136−4x)R=(100+10x)(136-4x) gives x=12x=12 blocks, i.e. 220 passengers.

Revenue R=(number of passengers)×(fare per passenger)R=(\text{number of passengers})\times(\text{fare per passenger}). Maximise via R′(x)=0R'(x)=0, R′′(x)<0R''(x)<0.

  1. Let xx be the number of complete groups of 10 passengers in excess of 100.
  2. Number of passengers =100+10x=100+10x; the fare falls ₹4 per group, so fare per passenger =₹(136−4x)=₹(136-4x).
  3. Revenue: R(x)=(100+10x)(136−4x)R(x)=(100+10x)(136-4x).
  4. Expand: R(x)=13600−400x+1360x−40x2=13600+960x−40x2R(x)=13600-400x+1360x-40x^2=13600+960x-40x^2.
  5. Differentiate: R′(x)=960−80xR'(x)=960-80x; set R′(x)=0⇒x=12R'(x)=0\Rightarrow x=12. …

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