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Ex 6.5, 15 - Find x and y, sum is 35 and x2 y5 is maximum

Ex 6.5,15 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.5,15 - Chapter 6 Class 12 Application of Derivatives - Part 3
Ex 6.5,15 - Chapter 6 Class 12 Application of Derivatives - Part 4
Ex 6.5,15 - Chapter 6 Class 12 Application of Derivatives - Part 5
Ex 6.5,15 - Chapter 6 Class 12 Application of Derivatives - Part 6
Ex 6.5,15 - Chapter 6 Class 12 Application of Derivatives - Part 7
Ex 6.5,15 - Chapter 6 Class 12 Application of Derivatives - Part 8
Ex 6.5,15 - Chapter 6 Class 12 Application of Derivatives - Part 9
Ex 6.5,15 - Chapter 6 Class 12 Application of Derivatives - Part 10
Ex 6.5,15 - Chapter 6 Class 12 Application of Derivatives - Part 11


Transcript

Ex 6.5, 15 (Method 1) Find two positive numbers π‘₯ and 𝑦 such that their sum is 35 and the product π‘₯2 𝑦5 is a maximum. Given two number are π‘₯ & 𝑦 Such that π‘₯ + 𝑦 = 35 𝑦 = 35 – π‘₯ Let P = π‘₯2 𝑦5 We need to maximize P Finding P’(𝒙) P(π‘₯)=π‘₯^2 𝑦^5 P(π‘₯)=π‘₯^2 (35βˆ’π‘₯)^5 P’(π‘₯)=𝑑(π‘₯^2 (35 βˆ’ π‘₯)^5 )/𝑑π‘₯ P’(π‘₯)=𝑑(π‘₯^2 )/𝑑π‘₯ . (35βˆ’π‘₯)^5+(𝑑(35 βˆ’ π‘₯)^5)/𝑑π‘₯ . π‘₯^2 =2π‘₯ .(35βˆ’π‘₯)^5+γ€–5(35βˆ’π‘₯)γ€—^4 .𝑑(35 βˆ’ π‘₯)/𝑑π‘₯ . π‘₯^2 =2π‘₯ .(35βˆ’π‘₯)^5+γ€–5(35βˆ’π‘₯)γ€—^4 . (0βˆ’1)(π‘₯^2 ) =2π‘₯ .(35βˆ’π‘₯)^5+γ€–5(35βˆ’π‘₯)γ€—^4 (βˆ’π‘₯^2 ) =2π‘₯ (35βˆ’π‘₯)^5βˆ’γ€–5π‘₯^2 (35βˆ’π‘₯)γ€—^4 = γ€– π‘₯ (35βˆ’π‘₯)γ€—^4 [2(35βˆ’π‘₯)βˆ’5π‘₯] = γ€– π‘₯ (35βˆ’π‘₯)γ€—^4 (70βˆ’7π‘₯) using product rule as (𝑒𝑣)^β€²=𝑒^β€² 𝑣+𝑣^β€² 𝑒 Putting P’(𝒙)=𝟎 γ€– π‘₯ (35βˆ’π‘₯)γ€—^4 (70βˆ’7π‘₯)=0 Hence π‘₯ = 0 , 10 , 35 are Critical Points But, If we Take π‘₯ = 0 Product will be 0 So, x = 0 is not possible (35βˆ’π‘₯)^4= 0 35βˆ’π‘₯=0 π‘₯=35 70βˆ’7π‘₯= 0 7π‘₯=70 π‘₯=70/7 π‘₯= 10 If x = 35 𝑦 = 35 – 35 = 35 – 35 = 0 So, product will be 0 So, x = 35 is not possible Hence only critical point is π‘₯=10 Finding P’’(𝒙) P’(π‘₯)=π‘₯(35βˆ’π‘₯)^4 (70βˆ’7π‘₯) P’(π‘₯)=(35βˆ’π‘₯)^4 (70π‘₯βˆ’7π‘₯^2 ) P’’(π‘₯)=(𝑑(35 βˆ’ π‘₯)^4)/𝑑π‘₯. (70π‘₯βˆ’7π‘₯^2 )+𝑑(70π‘₯ βˆ’ 7π‘₯^2 )/𝑑π‘₯ (35βˆ’π‘₯)^4 =4(35βˆ’π‘₯)^3.𝑑(35 βˆ’ π‘₯)/𝑑π‘₯. (70π‘₯βˆ’7π‘₯^2 )+(70βˆ’14π‘₯) (35βˆ’π‘₯)^4 Using product rule as (𝑒𝑣)^β€²=𝑒^β€² 𝑣+𝑣^β€² 𝑒 =4(35βˆ’π‘₯)^3 (0βˆ’1)(70π‘₯βˆ’7π‘₯^2 )+(70βˆ’14π‘₯) (35βˆ’π‘₯)^4 =βˆ’4(35βˆ’π‘₯)^3 (70π‘₯βˆ’7π‘₯^2 )+(70βˆ’14π‘₯) (35βˆ’π‘₯)^4 Putting π‘₯ = 10 in P’’(x) P’’(π‘₯) = βˆ’4(35βˆ’π‘₯)^3 (70π‘₯βˆ’7π‘₯^2 )+(70βˆ’14π‘₯) (35βˆ’π‘₯)^4 =βˆ’4(35βˆ’10)^3 (70(10)βˆ’7(10)^2 )+(70βˆ’14(10)) (35βˆ’10)^4 =βˆ’4(25)^3 (700βˆ’700)+(70βˆ’140) (25)^4 =βˆ’4(25)^3 (0)+(βˆ’70) (25)^4 =0βˆ’70(25)^4 =βˆ’70(25)^4 < 0 Thus, P’’(π‘₯)<0 when π‘₯ = 10 ∴ P is maximum when π‘₯ = 10 Thus, when π‘₯ = 10 𝑦 = 35 – π‘₯= 35 βˆ’10=25 Hence 𝒙 = 10 & π’š = 25 Ex 6.5, 15 (Method 2) Find two positive numbers π‘₯ and 𝑦 such that their sum is 35 and the product π‘₯2 𝑦5 is a maximum. Given two number are π‘₯ & 𝑦 Such that π‘₯ + 𝑦 = 35 𝑦 = 35 – π‘₯ Let P = π‘₯2 𝑦5 We need to maximise P Finding P’(𝒙) P(π‘₯)=π‘₯^2 𝑦^5 P(π‘₯)=π‘₯^2 (35βˆ’π‘₯)^5 P’(π‘₯)=𝑑(π‘₯^2 (35 βˆ’ π‘₯)^5 )/𝑑π‘₯ P’(π‘₯)=𝑑(π‘₯^2 )/𝑑π‘₯ . (35βˆ’π‘₯)^5+(𝑑(35 βˆ’ π‘₯)^5)/𝑑π‘₯ . π‘₯^2 =2π‘₯ .(35βˆ’π‘₯)^5+γ€–5(35βˆ’π‘₯)γ€—^4 .𝑑(35 βˆ’ π‘₯)/𝑑π‘₯ . π‘₯^2 =2π‘₯ .(35βˆ’π‘₯)^5+γ€–5(35βˆ’π‘₯)γ€—^4 . (0βˆ’1)(π‘₯^2 ) =2π‘₯ .(35βˆ’π‘₯)^5+γ€–5(35βˆ’π‘₯)γ€—^4 (βˆ’π‘₯^2 ) =2π‘₯ (35βˆ’π‘₯)^5βˆ’γ€–5π‘₯^2 (35βˆ’π‘₯)γ€—^4 = γ€– π‘₯ (35βˆ’π‘₯)γ€—^4 [2(35βˆ’π‘₯)βˆ’5π‘₯] = γ€– π‘₯ (35βˆ’π‘₯)γ€—^4 (70βˆ’7π‘₯) Using product rule as (𝑒𝑣)^β€²=𝑒^β€² 𝑣+𝑣^β€² 𝑒 Putting P’(𝒙)=𝟎 γ€–π‘₯ (35βˆ’π‘₯)γ€—^4 (70βˆ’7π‘₯)=0 γ€–π‘₯ (35βˆ’π‘₯)γ€—^4 (70βˆ’7π‘₯)=0 Hence π‘₯ = 0 , 10 , 35 are Critical Points But, If We Take π‘₯ = 0 Product will be 0 So, x = 0 is not possible (35βˆ’π‘₯)^4= 0 35βˆ’π‘₯=0 π‘₯=35 70βˆ’7π‘₯= 0 7π‘₯=70 π‘₯=70/7 π‘₯= 10 If x = 35 𝑦 = 35 – 35 = 35 – 35 = 0 So, product will be 0 So, x = 35 is not possible Hence only critical point is π‘₯=10 ∴ π‘₯ = 10 is point of maxima P(π‘₯) is maximum at π‘₯ = 10 Thus, when π‘₯ = 10 𝑦 = 35 – π‘₯= 35 βˆ’10=25 Hence 𝒙 = 10 & π’š = 25

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.