Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12










Last updated at Jan. 7, 2020 by Teachoo
Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12
Transcript
Ex 6.5, 15 (Method 1) Find two positive numbers ๐ฅ and ๐ฆ such that their sum is 35 and the product ๐ฅ2 ๐ฆ5 is a maximum. Given two number are ๐ฅ & ๐ฆ Such that ๐ฅ + ๐ฆ = 35 ๐ฆ = 35 โ ๐ฅ Let P = ๐ฅ2 ๐ฆ5 We need to maximize P Finding Pโ(๐) P(๐ฅ)=๐ฅ^2 ๐ฆ^5 P(๐ฅ)=๐ฅ^2 (35โ๐ฅ)^5 โฆ(1) Pโ(๐ฅ)=๐(๐ฅ^2 (35 โ ๐ฅ)^5 )/๐๐ฅ Pโ(๐ฅ)=๐(๐ฅ^2 )/๐๐ฅ . (35โ๐ฅ)^5+(๐(35 โ ๐ฅ)^5)/๐๐ฅ . ๐ฅ^2 =2๐ฅ .(35โ๐ฅ)^5+ใ5(35โ๐ฅ)ใ^4 .๐(35 โ ๐ฅ)/๐๐ฅ . ๐ฅ^2 =2๐ฅ .(35โ๐ฅ)^5+ใ5(35โ๐ฅ)ใ^4 . (0โ1)(๐ฅ^2 ) =2๐ฅ .(35โ๐ฅ)^5+ใ5(35โ๐ฅ)ใ^4 (โ๐ฅ^2 ) =2๐ฅ (35โ๐ฅ)^5โใ5๐ฅ^2 (35โ๐ฅ)ใ^4 = ใ ๐ฅ (35โ๐ฅ)ใ^4 [2(35โ๐ฅ)โ5๐ฅ] = ใ ๐ฅ (35โ๐ฅ)ใ^4 (70โ7๐ฅ) using product rule as (๐ข๐ฃ)^โฒ=๐ข^โฒ ๐ฃ+๐ฃ^โฒ ๐ข Putting Pโ(๐)=๐ ใ ๐ฅ (35โ๐ฅ)ใ^4 (70โ7๐ฅ)=0 Hence ๐ฅ = 0 , 10 , 35 are Critical Points But, If we Take ๐ฅ = 0 Product will be 0 So, x = 0 is not possible ๐ฅ = 0 (35โ๐ฅ)^4= 0 35โ๐ฅ=0 ๐ฅ=35 70โ7๐ฅ= 0 7๐ฅ=70 ๐ฅ=70/7 ๐ฅ= 10 If x = 35 ๐ฆ = 35 โ 35 = 35 โ 35 = 0 So, product will be 0 So, x = 35 is not possible Hence only critical point is ๐ฅ=10 Finding Pโโ(๐) Pโ(๐ฅ)=๐ฅ(35โ๐ฅ)^4 (70โ7๐ฅ) Pโ(๐ฅ)=(35โ๐ฅ)^4 (70๐ฅโ7๐ฅ^2 ) Pโโ(๐ฅ)=(๐(35 โ ๐ฅ)^4)/๐๐ฅ. (70๐ฅโ7๐ฅ^2 )+๐(70๐ฅ โ 7๐ฅ^2 )/๐๐ฅ (35โ๐ฅ)^4 =4(35โ๐ฅ)^3.๐(35 โ ๐ฅ)/๐๐ฅ. (70๐ฅโ7๐ฅ^2 )+(70โ14๐ฅ) (35โ๐ฅ)^4 Using product rule as (๐ข๐ฃ)^โฒ=๐ข^โฒ ๐ฃ+๐ฃ^โฒ ๐ข =4(35โ๐ฅ)^3 (0โ1)(70๐ฅโ7๐ฅ^2 )+(70โ14๐ฅ) (35โ๐ฅ)^4 =โ4(35โ๐ฅ)^3 (70๐ฅโ7๐ฅ^2 )+(70โ14๐ฅ) (35โ๐ฅ)^4 Putting ๐ฅ = 10 in Pโโ(x) Pโโ(๐ฅ) = โ4(35โ๐ฅ)^3 (70๐ฅโ7๐ฅ^2 )+(70โ14๐ฅ) (35โ๐ฅ)^4 =โ4(35โ10)^3 (70(10)โ7(10)^2 )+(70โ14(10)) (35โ10)^4 =โ4(25)^3 (700โ700)+(70โ140) (25)^4 =โ4(25)^3 (0)+(โ70) (25)^4 =0โ70(25)^4 =โ70(25)^4 < 0 Thus, Pโโ(๐ฅ)<0 when ๐ฅ = 10 โด P is maximum when ๐ฅ = 10 Thus, when ๐ฅ = 10 ๐ฆ = 35 โ ๐ฅ= 35 โ10=25 Hence ๐ = 10 & ๐ = 25 Ex 6.5, 15 (Method 2) Find two positive numbers ๐ฅ and ๐ฆ such that their sum is 35 and the product ๐ฅ2 ๐ฆ5 is a maximum. Given two number are ๐ฅ & ๐ฆ Such that ๐ฅ + ๐ฆ = 35 ๐ฆ = 35 โ ๐ฅ Let P = ๐ฅ2 ๐ฆ5 We need to maximise P Finding Pโ(๐) P(๐ฅ)=๐ฅ^2 ๐ฆ^5 P(๐ฅ)=๐ฅ^2 (35โ๐ฅ)^5 Pโ(๐ฅ)=๐(๐ฅ^2 (35 โ ๐ฅ)^5 )/๐๐ฅ Pโ(๐ฅ)=๐(๐ฅ^2 )/๐๐ฅ . (35โ๐ฅ)^5+(๐(35 โ ๐ฅ)^5)/๐๐ฅ . ๐ฅ^2 =2๐ฅ .(35โ๐ฅ)^5+ใ5(35โ๐ฅ)ใ^4 .๐(35 โ ๐ฅ)/๐๐ฅ . ๐ฅ^2 =2๐ฅ .(35โ๐ฅ)^5+ใ5(35โ๐ฅ)ใ^4 . (0โ1)(๐ฅ^2 ) =2๐ฅ .(35โ๐ฅ)^5+ใ5(35โ๐ฅ)ใ^4 (โ๐ฅ^2 ) =2๐ฅ (35โ๐ฅ)^5โใ5๐ฅ^2 (35โ๐ฅ)ใ^4 = ใ ๐ฅ (35โ๐ฅ)ใ^4 [2(35โ๐ฅ)โ5๐ฅ] = ใ ๐ฅ (35โ๐ฅ)ใ^4 (70โ7๐ฅ) Using product rule as (๐ข๐ฃ)^โฒ=๐ข^โฒ ๐ฃ+๐ฃ^โฒ ๐ข Putting Pโ(๐)=๐ ใ๐ฅ (35โ๐ฅ)ใ^4 (70โ7๐ฅ)=0 ใ๐ฅ (35โ๐ฅ)ใ^4 (70โ7๐ฅ)=0 Hence ๐ฅ = 0 , 10 , 35 are Critical Points But, If We Take ๐ฅ = 0 Product will be 0 So, x = 0 is not possible ๐ฅ = 0 (35โ๐ฅ)^4= 0 35โ๐ฅ=0 ๐ฅ=35 70โ7๐ฅ= 0 7๐ฅ=70 ๐ฅ=70/7 ๐ฅ= 10 If x = 35 ๐ฆ = 35 โ 35 = 35 โ 35 = 0 So, product will be 0 So, x = 35 is not possible Hence only critical point is ๐ฅ=10 โด ๐ฅ = 10 is point of maxima P(๐ฅ) is maximum at ๐ฅ = 10 Thus, when ๐ฅ = 10 ๐ฆ = 35 โ ๐ฅ= 35 โ10=25 Hence ๐ = 10 & ๐ = 25
Ex 6.5
Ex 6.5,2 Important
Ex 6.5,3
Ex 6.5,4
Ex 6.5,5 Important
Ex 6.5,6
Ex 6.5,7 Important
Ex 6.5,8
Ex 6.5,9 Important
Ex 6.5,10
Ex 6.5,11 Important
Ex 6.5,12 Important
Ex 6.5,13
Ex 6.5,14 Important
Ex 6.5,15 Important You are here
Ex 6.5,16
Ex 6.5,17
Ex 6.5,18 Important
Ex 6.5,19 Important
Ex 6.5, 20 Important
Ex 6.5,21
Ex 6.5,22 Important
Ex 6.5,23 Important
Ex 6.5,24 Important
Ex 6.5,25 Important
Ex 6.5,26 Important
Ex 6.5, 27
Ex 6.5,28 Important
Ex 6.5,29
About the Author