Ex 6.3

Chapter 6 Class 12 Application of Derivatives
Serial order wise

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Ex 6.3, 14 (Method 1) Find two positive numbers π₯ and y such that π₯ + π¦ = 60 and π₯π¦3 is maximum. Given two number π₯ and y, such that π₯ + π¦ = 60 π¦=60βπ₯ Let P = π₯π¦3 We need to maximize P Now, P = π₯π¦3 Putting value of y from (1) P = π₯(60βπ₯)3 Finding Pβ(x) P = π₯(60βπ₯)^3 Diff w.r.t π₯ ππ/ππ₯=π(π₯(60 β π₯)^3 )/ππ₯ ππ/ππ₯=π(π₯)/ππ₯ (60βπ₯)^3+(π(60 β π₯)^3)/ππ₯ . π₯ =(60βπ₯)^3+γ3(60βπ₯)γ^2 . (0β1) . π₯ =(60βπ₯)^3β3π₯(60βπ₯)^2 =(60βπ₯)^2 (60βπ₯)β3π₯(60βπ₯)^2 =(60βπ₯)^2 [(60βπ₯)β3π₯] =(60βπ₯)^2 [60β4π₯] Finding Pβ(x) P = π₯(60βπ₯)^3 Diff w.r.t π₯ ππ/ππ₯=π(π₯(60 β π₯)^3 )/ππ₯ ππ/ππ₯=π(π₯)/ππ₯ (60βπ₯)^3+(π(60 β π₯)^3)/ππ₯ . π₯ =(60βπ₯)^3+γ3(60βπ₯)γ^2 . (0β1) . π₯ =(60βπ₯)^3β3π₯(60βπ₯)^2 =(60βπ₯)^2 (60βπ₯)β3π₯(60βπ₯)^2 =(60βπ₯)^2 [(60βπ₯)β3π₯] =(60βπ₯)^2 [60β4π₯] Using product rule as (π’π£)^β²=π’^β² π£+π£^β² π’ Putting ππ·/ππ=π (60βπ₯)^2 (60β4π₯)=0 So, x = 60 & x = 60/4 = 15 But, If π₯=60, π¦= 60 β π₯ = 60 β 60 = 0 Which is not possible Hence, π₯= 15 is only critical point. Finding Pββ (π) Pββ (π₯)=π((60 β π₯)^2 (60 β 4π₯))/ππ₯ Pββ (π₯)=(π(60 β π₯)^2)/ππ₯ . (60β4π₯)+π(60 β 4π₯)/ππ₯ (60βπ₯)^2 = 2(60βπ₯) .(0β1)(60β4π₯)β4(60βπ₯)^2 = β2(60βπ₯) . (60β4π₯)β4(60βπ₯)^2 = β2(60βπ₯)[(60β4π₯)+2(60βπ₯)] = β2(60βπ₯)[(60β4π₯)+120β2π₯] = β2(60βπ₯)(180β6π₯) Using product rule as (π’π£)^β²=π’^β² π£+π£^β² π’ At π = 15 Pββ(15)=β2(60β15)(180β6(15)) =β90 Γ90 =β8100 < 0 β΄ Pββ(π₯)<0 at π₯ = 15 Hence π₯π¦3 is Maximum when π₯ = 15 Thus, when π₯ = 15 π¦ =60 β π₯=60 β15=45 Hence, numbers are 15 & 45 Ex 6.3, 14 (Method 2) Find two positive numbers π₯ and y such that π₯ + π¦ = 60 and π₯π¦3 is maximum. Given two number π₯ and y, such that π₯ + π¦ = 60 π¦=60βπ₯ Let P = π₯π¦3 We need to maximize P Now, P = π₯π¦3 P = π₯(60βπ₯)3 Finding Pβ(x) P = π₯(60βπ₯)^3 Diff w.r.t π₯ ππ/ππ₯=π(π₯(60βπ₯)^3 )/ππ₯ ππ/ππ₯=π(π₯)/ππ₯ (60βπ₯)^3+(π(60βπ₯)^3)/ππ₯ . π₯ =(60βπ₯)^3+γ3(60βπ₯)γ^2 . (0β1) . π₯ =(60βπ₯)^3+3π₯(60βπ₯)^2 =(60βπ₯)^2 [(60βπ₯)β3π₯] =(60βπ₯)^2 [60β4π₯] Using product rule as (π’π£)^β²=π’^β² π£+π£^β² π’ Putting ππ·/ππ=π (60βπ₯)^2 (60β4π₯)=0 So, x = 60 & x = 60/4 = 15 But, If π₯=60 π¦= 60 β 60 = 0 Which is not possible Hence π₯= 15 is only critical point. β΄ π₯ = 15 is point of local Maxima & P(π₯) is Maximum at π₯ = 15 Thus, when π₯ = 15 π¦ =60 β π₯=60 β15=45 Hence, numbers are 15 & 45