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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


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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.