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

Transcript

Ex 6.5, 15 (Method 1) Find two positive numbers ๐‘ฅ and ๐‘ฆ such that their sum is 35 and the product ๐‘ฅ2 ๐‘ฆ5 is a maximum. Given two number are ๐‘ฅ & ๐‘ฆ Such that ๐‘ฅ + ๐‘ฆ = 35 ๐‘ฆ = 35 โ€“ ๐‘ฅ Let P = ๐‘ฅ2 ๐‘ฆ5 We need to maximize P Finding Pโ€™(๐’™) P(๐‘ฅ)=๐‘ฅ^2 ๐‘ฆ^5 P(๐‘ฅ)=๐‘ฅ^2 (35โˆ’๐‘ฅ)^5 โ€ฆ(1) Pโ€™(๐‘ฅ)=๐‘‘(๐‘ฅ^2 (35 โˆ’ ๐‘ฅ)^5 )/๐‘‘๐‘ฅ Pโ€™(๐‘ฅ)=๐‘‘(๐‘ฅ^2 )/๐‘‘๐‘ฅ . (35โˆ’๐‘ฅ)^5+(๐‘‘(35 โˆ’ ๐‘ฅ)^5)/๐‘‘๐‘ฅ . ๐‘ฅ^2 =2๐‘ฅ .(35โˆ’๐‘ฅ)^5+ใ€–5(35โˆ’๐‘ฅ)ใ€—^4 .๐‘‘(35 โˆ’ ๐‘ฅ)/๐‘‘๐‘ฅ . ๐‘ฅ^2 =2๐‘ฅ .(35โˆ’๐‘ฅ)^5+ใ€–5(35โˆ’๐‘ฅ)ใ€—^4 . (0โˆ’1)(๐‘ฅ^2 ) =2๐‘ฅ .(35โˆ’๐‘ฅ)^5+ใ€–5(35โˆ’๐‘ฅ)ใ€—^4 (โˆ’๐‘ฅ^2 ) =2๐‘ฅ (35โˆ’๐‘ฅ)^5โˆ’ใ€–5๐‘ฅ^2 (35โˆ’๐‘ฅ)ใ€—^4 = ใ€– ๐‘ฅ (35โˆ’๐‘ฅ)ใ€—^4 [2(35โˆ’๐‘ฅ)โˆ’5๐‘ฅ] = ใ€– ๐‘ฅ (35โˆ’๐‘ฅ)ใ€—^4 (70โˆ’7๐‘ฅ) using product rule as (๐‘ข๐‘ฃ)^โ€ฒ=๐‘ข^โ€ฒ ๐‘ฃ+๐‘ฃ^โ€ฒ ๐‘ข Putting Pโ€™(๐’™)=๐ŸŽ ใ€– ๐‘ฅ (35โˆ’๐‘ฅ)ใ€—^4 (70โˆ’7๐‘ฅ)=0 Hence ๐‘ฅ = 0 , 10 , 35 are Critical Points But, If we Take ๐‘ฅ = 0 Product will be 0 So, x = 0 is not possible ๐‘ฅ = 0 (35โˆ’๐‘ฅ)^4= 0 35โˆ’๐‘ฅ=0 ๐‘ฅ=35 70โˆ’7๐‘ฅ= 0 7๐‘ฅ=70 ๐‘ฅ=70/7 ๐‘ฅ= 10 If x = 35 ๐‘ฆ = 35 โ€“ 35 = 35 โ€“ 35 = 0 So, product will be 0 So, x = 35 is not possible Hence only critical point is ๐‘ฅ=10 Finding Pโ€™โ€™(๐’™) Pโ€™(๐‘ฅ)=๐‘ฅ(35โˆ’๐‘ฅ)^4 (70โˆ’7๐‘ฅ) Pโ€™(๐‘ฅ)=(35โˆ’๐‘ฅ)^4 (70๐‘ฅโˆ’7๐‘ฅ^2 ) Pโ€™โ€™(๐‘ฅ)=(๐‘‘(35 โˆ’ ๐‘ฅ)^4)/๐‘‘๐‘ฅ. (70๐‘ฅโˆ’7๐‘ฅ^2 )+๐‘‘(70๐‘ฅ โˆ’ 7๐‘ฅ^2 )/๐‘‘๐‘ฅ (35โˆ’๐‘ฅ)^4 =4(35โˆ’๐‘ฅ)^3.๐‘‘(35 โˆ’ ๐‘ฅ)/๐‘‘๐‘ฅ. (70๐‘ฅโˆ’7๐‘ฅ^2 )+(70โˆ’14๐‘ฅ) (35โˆ’๐‘ฅ)^4 Using product rule as (๐‘ข๐‘ฃ)^โ€ฒ=๐‘ข^โ€ฒ ๐‘ฃ+๐‘ฃ^โ€ฒ ๐‘ข =4(35โˆ’๐‘ฅ)^3 (0โˆ’1)(70๐‘ฅโˆ’7๐‘ฅ^2 )+(70โˆ’14๐‘ฅ) (35โˆ’๐‘ฅ)^4 =โˆ’4(35โˆ’๐‘ฅ)^3 (70๐‘ฅโˆ’7๐‘ฅ^2 )+(70โˆ’14๐‘ฅ) (35โˆ’๐‘ฅ)^4 Putting ๐‘ฅ = 10 in Pโ€™โ€™(x) Pโ€™โ€™(๐‘ฅ) = โˆ’4(35โˆ’๐‘ฅ)^3 (70๐‘ฅโˆ’7๐‘ฅ^2 )+(70โˆ’14๐‘ฅ) (35โˆ’๐‘ฅ)^4 =โˆ’4(35โˆ’10)^3 (70(10)โˆ’7(10)^2 )+(70โˆ’14(10)) (35โˆ’10)^4 =โˆ’4(25)^3 (700โˆ’700)+(70โˆ’140) (25)^4 =โˆ’4(25)^3 (0)+(โˆ’70) (25)^4 =0โˆ’70(25)^4 =โˆ’70(25)^4 < 0 Thus, Pโ€™โ€™(๐‘ฅ)<0 when ๐‘ฅ = 10 โˆด P is maximum when ๐‘ฅ = 10 Thus, when ๐‘ฅ = 10 ๐‘ฆ = 35 โ€“ ๐‘ฅ= 35 โˆ’10=25 Hence ๐’™ = 10 & ๐’š = 25 Ex 6.5, 15 (Method 2) Find two positive numbers ๐‘ฅ and ๐‘ฆ such that their sum is 35 and the product ๐‘ฅ2 ๐‘ฆ5 is a maximum. Given two number are ๐‘ฅ & ๐‘ฆ Such that ๐‘ฅ + ๐‘ฆ = 35 ๐‘ฆ = 35 โ€“ ๐‘ฅ Let P = ๐‘ฅ2 ๐‘ฆ5 We need to maximise P Finding Pโ€™(๐’™) P(๐‘ฅ)=๐‘ฅ^2 ๐‘ฆ^5 P(๐‘ฅ)=๐‘ฅ^2 (35โˆ’๐‘ฅ)^5 Pโ€™(๐‘ฅ)=๐‘‘(๐‘ฅ^2 (35 โˆ’ ๐‘ฅ)^5 )/๐‘‘๐‘ฅ Pโ€™(๐‘ฅ)=๐‘‘(๐‘ฅ^2 )/๐‘‘๐‘ฅ . (35โˆ’๐‘ฅ)^5+(๐‘‘(35 โˆ’ ๐‘ฅ)^5)/๐‘‘๐‘ฅ . ๐‘ฅ^2 =2๐‘ฅ .(35โˆ’๐‘ฅ)^5+ใ€–5(35โˆ’๐‘ฅ)ใ€—^4 .๐‘‘(35 โˆ’ ๐‘ฅ)/๐‘‘๐‘ฅ . ๐‘ฅ^2 =2๐‘ฅ .(35โˆ’๐‘ฅ)^5+ใ€–5(35โˆ’๐‘ฅ)ใ€—^4 . (0โˆ’1)(๐‘ฅ^2 ) =2๐‘ฅ .(35โˆ’๐‘ฅ)^5+ใ€–5(35โˆ’๐‘ฅ)ใ€—^4 (โˆ’๐‘ฅ^2 ) =2๐‘ฅ (35โˆ’๐‘ฅ)^5โˆ’ใ€–5๐‘ฅ^2 (35โˆ’๐‘ฅ)ใ€—^4 = ใ€– ๐‘ฅ (35โˆ’๐‘ฅ)ใ€—^4 [2(35โˆ’๐‘ฅ)โˆ’5๐‘ฅ] = ใ€– ๐‘ฅ (35โˆ’๐‘ฅ)ใ€—^4 (70โˆ’7๐‘ฅ) Using product rule as (๐‘ข๐‘ฃ)^โ€ฒ=๐‘ข^โ€ฒ ๐‘ฃ+๐‘ฃ^โ€ฒ ๐‘ข Putting Pโ€™(๐’™)=๐ŸŽ ใ€–๐‘ฅ (35โˆ’๐‘ฅ)ใ€—^4 (70โˆ’7๐‘ฅ)=0 ใ€–๐‘ฅ (35โˆ’๐‘ฅ)ใ€—^4 (70โˆ’7๐‘ฅ)=0 Hence ๐‘ฅ = 0 , 10 , 35 are Critical Points But, If We Take ๐‘ฅ = 0 Product will be 0 So, x = 0 is not possible ๐‘ฅ = 0 (35โˆ’๐‘ฅ)^4= 0 35โˆ’๐‘ฅ=0 ๐‘ฅ=35 70โˆ’7๐‘ฅ= 0 7๐‘ฅ=70 ๐‘ฅ=70/7 ๐‘ฅ= 10 If x = 35 ๐‘ฆ = 35 โ€“ 35 = 35 โ€“ 35 = 0 So, product will be 0 So, x = 35 is not possible Hence only critical point is ๐‘ฅ=10 โˆด ๐‘ฅ = 10 is point of maxima P(๐‘ฅ) is maximum at ๐‘ฅ = 10 Thus, when ๐‘ฅ = 10 ๐‘ฆ = 35 โ€“ ๐‘ฅ= 35 โˆ’10=25 Hence ๐’™ = 10 & ๐’š = 25

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.