Ex 6.5, 6 - Find maximum profit that a company can make, p(x)

Ex 6.5,6 - Chapter 6 Class 12 Application of Derivatives - Part 2

  1. Chapter 6 Class 12 Application of Derivatives (Term 1)
  2. Serial order wise

Transcript

Ex 6.5, 6 Find the maximum profit that a company can make, if the profit function is given by ๐‘(๐‘ฅ) = 41 โ€“ 72๐‘ฅ โ€“ 18๐‘ฅ2The profit function is given by p(x) = 41 โˆ’ 72x โˆ’ 18x2 pโ€™(x) = โˆ’72 โˆ’ 36 x Putting pโ€˜ (x) = 0 โˆ’72 โˆ’ 36x = 0 โˆ’36x = 72 ๐‘ฅ =(โˆ’72)/36= โˆ’2 Now, Pโ€(x) = โˆ’36 Since Pโ€ (x) < 0 ๐‘ฅ=โˆ’2 is the maxima โˆด Maximum profit = p(โˆ’2) p(x) = 41 โˆ’ 72x โˆ’ 18x2 ๐‘(โˆ’2)=41โˆ’72 (โˆ’2)โˆ’18 (โˆ’2)^2 =41+144โˆ’18 (4) =41+144โˆ’72 =113 Hence, the maximum profit is 113 unit.

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.