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Ex 6.5, 14 - Find x and y such that x + y = 60, xy3 is max

Ex 6.5,14 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.5,14 - Chapter 6 Class 12 Application of Derivatives - Part 3 Ex 6.5,14 - Chapter 6 Class 12 Application of Derivatives - Part 4 Ex 6.5,14 - Chapter 6 Class 12 Application of Derivatives - Part 5 Ex 6.5,14 - Chapter 6 Class 12 Application of Derivatives - Part 6 Ex 6.5,14 - Chapter 6 Class 12 Application of Derivatives - Part 7 Ex 6.5,14 - Chapter 6 Class 12 Application of Derivatives - Part 8 Ex 6.5,14 - Chapter 6 Class 12 Application of Derivatives - Part 9

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Ex 6.5, 14 (Method 1) Find two positive numbers π‘₯ and y such that π‘₯ + 𝑦 = 60 and π‘₯𝑦3 is maximum. Given two number π‘₯ and y, such that π‘₯ + 𝑦 = 60 𝑦=60βˆ’π‘₯ Let P = π‘₯𝑦3 We need to maximize P Now, P = π‘₯𝑦3 Putting value of y from (1) P = π‘₯(60βˆ’π‘₯)3 Finding P’(x) P = π‘₯(60βˆ’π‘₯)^3 Diff w.r.t π‘₯ 𝑑𝑃/𝑑π‘₯=𝑑(π‘₯(60 βˆ’ π‘₯)^3 )/𝑑π‘₯ 𝑑𝑃/𝑑π‘₯=𝑑(π‘₯)/𝑑π‘₯ (60βˆ’π‘₯)^3+(𝑑(60 βˆ’ π‘₯)^3)/𝑑π‘₯ . π‘₯ =(60βˆ’π‘₯)^3+γ€–3(60βˆ’π‘₯)γ€—^2 . (0βˆ’1) . π‘₯ =(60βˆ’π‘₯)^3βˆ’3π‘₯(60βˆ’π‘₯)^2 =(60βˆ’π‘₯)^2 (60βˆ’π‘₯)βˆ’3π‘₯(60βˆ’π‘₯)^2 =(60βˆ’π‘₯)^2 [(60βˆ’π‘₯)βˆ’3π‘₯] =(60βˆ’π‘₯)^2 [60βˆ’4π‘₯] Finding P’(x) P = π‘₯(60βˆ’π‘₯)^3 Diff w.r.t π‘₯ 𝑑𝑃/𝑑π‘₯=𝑑(π‘₯(60 βˆ’ π‘₯)^3 )/𝑑π‘₯ 𝑑𝑃/𝑑π‘₯=𝑑(π‘₯)/𝑑π‘₯ (60βˆ’π‘₯)^3+(𝑑(60 βˆ’ π‘₯)^3)/𝑑π‘₯ . π‘₯ =(60βˆ’π‘₯)^3+γ€–3(60βˆ’π‘₯)γ€—^2 . (0βˆ’1) . π‘₯ =(60βˆ’π‘₯)^3βˆ’3π‘₯(60βˆ’π‘₯)^2 =(60βˆ’π‘₯)^2 (60βˆ’π‘₯)βˆ’3π‘₯(60βˆ’π‘₯)^2 =(60βˆ’π‘₯)^2 [(60βˆ’π‘₯)βˆ’3π‘₯] =(60βˆ’π‘₯)^2 [60βˆ’4π‘₯] Using product rule as (𝑒𝑣)^β€²=𝑒^β€² 𝑣+𝑣^β€² 𝑒 Putting 𝒅𝑷/𝒅𝒙=𝟎 (60βˆ’π‘₯)^2 (60βˆ’4π‘₯)=0 So, x = 60 & x = 60/4 = 15 But, If π‘₯=60, 𝑦= 60 – π‘₯ = 60 – 60 = 0 Which is not possible Hence, π‘₯= 15 is only critical point. Finding P’’ (𝒙) P’’ (π‘₯)=𝑑((60 βˆ’ π‘₯)^2 (60 βˆ’ 4π‘₯))/𝑑π‘₯ P’’ (π‘₯)=(𝑑(60 βˆ’ π‘₯)^2)/𝑑π‘₯ . (60βˆ’4π‘₯)+𝑑(60 βˆ’ 4π‘₯)/𝑑π‘₯ (60βˆ’π‘₯)^2 = 2(60βˆ’π‘₯) .(0βˆ’1)(60βˆ’4π‘₯)βˆ’4(60βˆ’π‘₯)^2 = βˆ’2(60βˆ’π‘₯) . (60βˆ’4π‘₯)βˆ’4(60βˆ’π‘₯)^2 = βˆ’2(60βˆ’π‘₯)[(60βˆ’4π‘₯)+2(60βˆ’π‘₯)] = βˆ’2(60βˆ’π‘₯)[(60βˆ’4π‘₯)+120βˆ’2π‘₯] = βˆ’2(60βˆ’π‘₯)(180βˆ’6π‘₯) Using product rule as (𝑒𝑣)^β€²=𝑒^β€² 𝑣+𝑣^β€² 𝑒 At 𝒙 = 15 P’’(15)=βˆ’2(60βˆ’15)(180βˆ’6(15)) =βˆ’90 Γ—90 =βˆ’8100 < 0 ∴ P’’(π‘₯)<0 at π‘₯ = 15 Hence π‘₯𝑦3 is Maximum when π‘₯ = 15 Thus, when π‘₯ = 15 𝑦 =60 – π‘₯=60 βˆ’15=45 Hence, numbers are 15 & 45 Ex 6.5, 14 (Method 2) Find two positive numbers π‘₯ and y such that π‘₯ + 𝑦 = 60 and π‘₯𝑦3 is maximum. Given two number π‘₯ and y, such that π‘₯ + 𝑦 = 60 𝑦=60βˆ’π‘₯ Let P = π‘₯𝑦3 We need to maximize P Now, P = π‘₯𝑦3 P = π‘₯(60βˆ’π‘₯)3 Finding P’(x) P = π‘₯(60βˆ’π‘₯)^3 Diff w.r.t π‘₯ 𝑑𝑃/𝑑π‘₯=𝑑(π‘₯(60βˆ’π‘₯)^3 )/𝑑π‘₯ 𝑑𝑃/𝑑π‘₯=𝑑(π‘₯)/𝑑π‘₯ (60βˆ’π‘₯)^3+(𝑑(60βˆ’π‘₯)^3)/𝑑π‘₯ . π‘₯ =(60βˆ’π‘₯)^3+γ€–3(60βˆ’π‘₯)γ€—^2 . (0βˆ’1) . π‘₯ =(60βˆ’π‘₯)^3+3π‘₯(60βˆ’π‘₯)^2 =(60βˆ’π‘₯)^2 [(60βˆ’π‘₯)βˆ’3π‘₯] =(60βˆ’π‘₯)^2 [60βˆ’4π‘₯] Using product rule as (𝑒𝑣)^β€²=𝑒^β€² 𝑣+𝑣^β€² 𝑒 Putting 𝒅𝑷/𝒅𝒙=𝟎 (60βˆ’π‘₯)^2 (60βˆ’4π‘₯)=0 So, x = 60 & x = 60/4 = 15 But, If π‘₯=60 𝑦= 60 – 60 = 0 Which is not possible Hence π‘₯= 15 is only critical point. ∴ π‘₯ = 15 is point of local Maxima & P(π‘₯) is Maximum at π‘₯ = 15 Thus, when π‘₯ = 15 𝑦 =60 – π‘₯=60 βˆ’15=45 Hence, numbers are 15 & 45

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Davneet Singh

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.