1. Chapter 6 Class 12 Application of Derivatives
2. Serial order wise
3. Ex 6.5

Transcript

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=60โ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(60โ๐ฅ)3 Step 1: Finding Pโ(x) P = ๐ฅ(60โ๐ฅ)^3 Diff w.r.t ๐ฅ ๐๐/๐๐ฅ=๐(๐ฅ(60 โ 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 (60โ๐ฅ)^3+(๐(60 โ ๐ฅ)^3)/๐๐ฅ . ๐ฅ =(60โ๐ฅ)^3+ใ3(60โ๐ฅ)ใ^2 . (0โ1) . ๐ฅ =(60โ๐ฅ)^3โ3๐ฅ(60โ๐ฅ)^2 =(60โ๐ฅ)^2 (60โ๐ฅ)โ3๐ฅ(60โ๐ฅ)^2 =(60โ๐ฅ)^2 [(60โ๐ฅ)โ3๐ฅ] =(60โ๐ฅ)^2 [60โ4๐ฅ] Step 2: Putting ๐๐/๐๐ฅ=0 (60โ๐ฅ)^2 (60โ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 โ ๐ฅ)^2 (60 โ 4๐ฅ))/๐๐ฅ Pโโ (๐ฅ)=(๐(60 โ ๐ฅ)^2)/๐๐ฅ . (60โ4๐ฅ)+๐(60 โ 4๐ฅ)/๐๐ฅ (60โ๐ฅ)^2 = 2(60โ๐ฅ) .(0โ1)(60โ4๐ฅ)โ4(60โ๐ฅ)^2 = โ2(60โ๐ฅ) . (60โ4๐ฅ)โ4(60โ๐ฅ)^2 = โ2(60โ๐ฅ)[(60โ4๐ฅ)+2(60โ๐ฅ)] = โ2(60โ๐ฅ)[(60โ4๐ฅ)+120โ2๐ฅ] = โ2(60โ๐ฅ)(180โ6๐ฅ) At ๐ฅ = 15 Pโโ(15)=โ2(60โ15)(180โ6(15)) =โ90 ร90 =โ8100 < 0 โด 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 โ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=60โ๐ฅ Let P = ๐ฅ๐ฆ3 We need to maximize P Now, P = ๐ฅ๐ฆ3 P = ๐ฅ(60โ๐ฅ)3 Step 1: Finding Pโ(x) P = ๐ฅ(60โ๐ฅ)^3 Diff w.r.t ๐ฅ ๐๐/๐๐ฅ=๐(๐ฅ(60โ๐ฅ)^3 )/๐๐ฅ ๐๐/๐๐ฅ=๐(๐ฅ)/๐๐ฅ (60โ๐ฅ)^3+(๐(60โ๐ฅ)^3)/๐๐ฅ . ๐ฅ =(60โ๐ฅ)^3+ใ3(60โ๐ฅ)ใ^2 . (0โ1) . ๐ฅ =(60โ๐ฅ)^3+3๐ฅ(60โ๐ฅ)^2 =(60โ๐ฅ)^2 [(60โ๐ฅ)โ3๐ฅ] =(60โ๐ฅ)^2 [60โ4๐ฅ] Step 2: Putting ๐๐/๐๐ฅ=0 (60โ๐ฅ)^2 (60โ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 โ15=45 Hence, numbers are 15 & 45

Ex 6.5