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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


Transcript

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 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.