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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

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Transcript

Ex 6.3, 6 Find the maximum profit that a company can make, if the profit function is given by 𝑝(π‘₯) = 41 – 72π‘₯ – 18π‘₯2The profit function is given by p(x) = 41 βˆ’ 72x βˆ’ 18x2 p’(x) = βˆ’72 βˆ’ 36 x Putting pβ€˜ (x) = 0 βˆ’72 βˆ’ 36x = 0 βˆ’36x = 72 π‘₯ =(βˆ’72)/36= βˆ’2 Now, P”(x) = βˆ’36 Since P” (x) < 0 π‘₯=βˆ’2 is the maxima ∴ Maximum profit = p(βˆ’2) p(x) = 41 βˆ’ 72x βˆ’ 18x2 𝑝(βˆ’2)=41βˆ’72 (βˆ’2)βˆ’18 (βˆ’2)^2 =41+144βˆ’18 (4) =41+144βˆ’72 =113 Hence, the maximum profit is 113 unit.

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.