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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise
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Ex 6.5,14 (Method 1) Find two positive numbers ๐‘ฅ and y such that ๐‘ฅ + ๐‘ฆ = 60 and ๐‘ฅ๐‘ฆ3 is maximum. Given two number ๐‘ฅ and y, such that ๐‘ฅ + ๐‘ฆ = 60 ๐‘ฆ=60โˆ’๐‘ฅ Let P = ๐‘ฅ๐‘ฆ3 We need to maximize P Now, P = ๐‘ฅ๐‘ฆ3 Putting value of y from (1) P = ๐‘ฅ(60โˆ’๐‘ฅ)3 Step 1: Finding Pโ€™(x) P = ๐‘ฅ(60โˆ’๐‘ฅ)^3 Diff w.r.t ๐‘ฅ ๐‘‘๐‘/๐‘‘๐‘ฅ=๐‘‘(๐‘ฅ(60 โˆ’ ๐‘ฅ)^3 )/๐‘‘๐‘ฅ Using product rule as (๐‘ข๐‘ฃ)^โ€ฒ=๐‘ข^โ€ฒ ๐‘ฃ+๐‘ฃ^โ€ฒ ๐‘ข ๐‘‘๐‘/๐‘‘๐‘ฅ=๐‘‘(๐‘ฅ)/๐‘‘๐‘ฅ (60โˆ’๐‘ฅ)^3+(๐‘‘(60 โˆ’ ๐‘ฅ)^3)/๐‘‘๐‘ฅ . ๐‘ฅ =(60โˆ’๐‘ฅ)^3+ใ€–3(60โˆ’๐‘ฅ)ใ€—^2 . (0โˆ’1) . ๐‘ฅ =(60โˆ’๐‘ฅ)^3โˆ’3๐‘ฅ(60โˆ’๐‘ฅ)^2 =(60โˆ’๐‘ฅ)^2 (60โˆ’๐‘ฅ)โˆ’3๐‘ฅ(60โˆ’๐‘ฅ)^2 =(60โˆ’๐‘ฅ)^2 [(60โˆ’๐‘ฅ)โˆ’3๐‘ฅ] =(60โˆ’๐‘ฅ)^2 [60โˆ’4๐‘ฅ] Step 2: Putting ๐‘‘๐‘ƒ/๐‘‘๐‘ฅ=0 (60โˆ’๐‘ฅ)^2 (60โˆ’4๐‘ฅ)=0 So, x = 60 & x = 60/4 = 15 But, If ๐‘ฅ=60 ๐‘ฆ= 60 โ€“ ๐‘ฅ = 60 โ€“ 60 = 0 Which is not possible Hence ๐‘ฅ= 15 is only critical point. Step 3: Finding Pโ€™โ€™ (๐‘ฅ) Pโ€™โ€™ (๐‘ฅ)=๐‘‘((60 โˆ’ ๐‘ฅ)^2 (60 โˆ’ 4๐‘ฅ))/๐‘‘๐‘ฅ Pโ€™โ€™ (๐‘ฅ)=(๐‘‘(60 โˆ’ ๐‘ฅ)^2)/๐‘‘๐‘ฅ . (60โˆ’4๐‘ฅ)+๐‘‘(60 โˆ’ 4๐‘ฅ)/๐‘‘๐‘ฅ (60โˆ’๐‘ฅ)^2 = 2(60โˆ’๐‘ฅ) .(0โˆ’1)(60โˆ’4๐‘ฅ)โˆ’4(60โˆ’๐‘ฅ)^2 = โˆ’2(60โˆ’๐‘ฅ) . (60โˆ’4๐‘ฅ)โˆ’4(60โˆ’๐‘ฅ)^2 = โˆ’2(60โˆ’๐‘ฅ)[(60โˆ’4๐‘ฅ)+2(60โˆ’๐‘ฅ)] = โˆ’2(60โˆ’๐‘ฅ)[(60โˆ’4๐‘ฅ)+120โˆ’2๐‘ฅ] = โˆ’2(60โˆ’๐‘ฅ)(180โˆ’6๐‘ฅ) At ๐‘ฅ = 15 Pโ€™โ€™(15)=โˆ’2(60โˆ’15)(180โˆ’6(15)) =โˆ’90 ร—90 =โˆ’8100 < 0 โˆด Pโ€™โ€™(๐‘ฅ)<0 at ๐‘ฅ = 15 Hence ๐‘ฅ๐‘ฆ3 is Maximum when ๐‘ฅ = 15 Thus, when ๐‘ฅ = 15 ๐‘ฆ =60 โ€“ ๐‘ฅ=60 โˆ’15=45 Hence, numbers are 15 & 45 Ex 6.5,14 (Method 2) Find two positive numbers ๐‘ฅ and y such that ๐‘ฅ + ๐‘ฆ = 60 and ๐‘ฅ๐‘ฆ3 is maximum. Given two number ๐‘ฅ and y, such that ๐‘ฅ + ๐‘ฆ = 60 ๐‘ฆ=60โˆ’๐‘ฅ Let P = ๐‘ฅ๐‘ฆ3 We need to maximize P Now, P = ๐‘ฅ๐‘ฆ3 P = ๐‘ฅ(60โˆ’๐‘ฅ)3 Step 1: Finding Pโ€™(x) P = ๐‘ฅ(60โˆ’๐‘ฅ)^3 Diff w.r.t ๐‘ฅ ๐‘‘๐‘/๐‘‘๐‘ฅ=๐‘‘(๐‘ฅ(60โˆ’๐‘ฅ)^3 )/๐‘‘๐‘ฅ ๐‘‘๐‘/๐‘‘๐‘ฅ=๐‘‘(๐‘ฅ)/๐‘‘๐‘ฅ (60โˆ’๐‘ฅ)^3+(๐‘‘(60โˆ’๐‘ฅ)^3)/๐‘‘๐‘ฅ . ๐‘ฅ =(60โˆ’๐‘ฅ)^3+ใ€–3(60โˆ’๐‘ฅ)ใ€—^2 . (0โˆ’1) . ๐‘ฅ =(60โˆ’๐‘ฅ)^3+3๐‘ฅ(60โˆ’๐‘ฅ)^2 =(60โˆ’๐‘ฅ)^2 [(60โˆ’๐‘ฅ)โˆ’3๐‘ฅ] =(60โˆ’๐‘ฅ)^2 [60โˆ’4๐‘ฅ] Step 2: Putting ๐‘‘๐‘ƒ/๐‘‘๐‘ฅ=0 (60โˆ’๐‘ฅ)^2 (60โˆ’4๐‘ฅ)=0 So, x = 60 & x = 60/4 = 15 But, If ๐‘ฅ=60 ๐‘ฆ= 60 โ€“ 60 = 0 Which is not possible Hence ๐‘ฅ= 15 is only critical point. Step 3: At ๐‘ฅ = 15 - if ๐‘ฅ < 15 (say 14.9) > 0 - if ๐‘ฅ > 15 (say 15.1) < 0 Since sign of Pโ€™(x) changes from positive to negative, it is Maxima โ‡’ ๐‘ฅ = 15 is point of local Maxima & P(๐‘ฅ) is Maximum at ๐‘ฅ = 15 Thus, when ๐‘ฅ = 15 ๐‘ฆ =60 โ€“ ๐‘ฅ=60 โˆ’15=45 Hence, numbers are 15 & 45

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.