







Ex 6.5
Ex 6.5, 1 (ii)
Ex 6.5, 1 (iii) Important
Ex 6.5, 1 (iv)
Ex 6.5, 2 (i)
Ex 6.5, 2 (ii) Important
Ex 6.5, 2 (iii)
Ex 6.5, 2 (iv) Important
Ex 6.5, 2 (v) Important
Ex 6.5, 3 (i)
Ex 6.5, 3 (ii)
Ex 6.5, 3 (iii)
Ex 6.5, 3 (iv) Important
Ex 6.5, 3 (v)
Ex 6.5, 3 (vi)
Ex 6.5, 3 (vii) Important
Ex 6.5, 3 (viii)
Ex 6.5, 4 (i)
Ex 6.5, 4 (ii) Important
Ex 6.5, 4 (iii)
Ex 6.5, 5 (i)
Ex 6.5, 5 (ii)
Ex 6.5, 5 (iii) Important
Ex 6.5, 5 (iv)
Ex 6.5,6
Ex 6.5,7 Important
Ex 6.5,8
Ex 6.5,9 Important
Ex 6.5,10
Ex 6.5,11 Important
Ex 6.5,12 Important
Ex 6.5,13
Ex 6.5,14 Important You are here
Ex 6.5,15 Important
Ex 6.5,16
Ex 6.5,17
Ex 6.5,18 Important
Ex 6.5,19 Important
Ex 6.5, 20 Important
Ex 6.5,21
Ex 6.5,22 Important
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Ex 6.5,24 Important
Ex 6.5,25 Important
Ex 6.5,26 Important
Ex 6.5, 27 (MCQ)
Ex 6.5,28 (MCQ) Important
Ex 6.5,29 (MCQ)
Last updated at April 15, 2021 by Teachoo
Ex 6.5, 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¦=60βπ₯ Let P = π₯π¦3 We need to maximize P Now, P = π₯π¦3 Putting value of y from (1) P = π₯(60βπ₯)3 Finding Pβ(x) P = π₯(60βπ₯)^3 Diff w.r.t π₯ ππ/ππ₯=π(π₯(60 β π₯)^3 )/ππ₯ ππ/ππ₯=π(π₯)/ππ₯ (60βπ₯)^3+(π(60 β π₯)^3)/ππ₯ . π₯ =(60βπ₯)^3+γ3(60βπ₯)γ^2 . (0β1) . π₯ =(60βπ₯)^3β3π₯(60βπ₯)^2 =(60βπ₯)^2 (60βπ₯)β3π₯(60βπ₯)^2 =(60βπ₯)^2 [(60βπ₯)β3π₯] =(60βπ₯)^2 [60β4π₯] Finding Pβ(x) P = π₯(60βπ₯)^3 Diff w.r.t π₯ ππ/ππ₯=π(π₯(60 β π₯)^3 )/ππ₯ ππ/ππ₯=π(π₯)/ππ₯ (60βπ₯)^3+(π(60 β π₯)^3)/ππ₯ . π₯ =(60βπ₯)^3+γ3(60βπ₯)γ^2 . (0β1) . π₯ =(60βπ₯)^3β3π₯(60βπ₯)^2 =(60βπ₯)^2 (60βπ₯)β3π₯(60βπ₯)^2 =(60βπ₯)^2 [(60βπ₯)β3π₯] =(60βπ₯)^2 [60β4π₯] Using product rule as (π’π£)^β²=π’^β² π£+π£^β² π’ Putting π π·/π π=π (60βπ₯)^2 (60β4π₯)=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 Pββ (π) Pββ (π₯)=π((60 β π₯)^2 (60 β 4π₯))/ππ₯ Pββ (π₯)=(π(60 β π₯)^2)/ππ₯ . (60β4π₯)+π(60 β 4π₯)/ππ₯ (60βπ₯)^2 = 2(60βπ₯) .(0β1)(60β4π₯)β4(60βπ₯)^2 = β2(60βπ₯) . (60β4π₯)β4(60βπ₯)^2 = β2(60βπ₯)[(60β4π₯)+2(60βπ₯)] = β2(60βπ₯)[(60β4π₯)+120β2π₯] = β2(60βπ₯)(180β6π₯) Using product rule as (π’π£)^β²=π’^β² π£+π£^β² π’ At π = 15 Pββ(15)=β2(60β15)(180β6(15)) =β90 Γ90 =β8100 < 0 β΄ Pββ(π₯)<0 at π₯ = 15 Hence π₯π¦3 is Maximum when π₯ = 15 Thus, when π₯ = 15 π¦ =60 β π₯=60 β15=45 Hence, numbers are 15 & 45 Ex 6.5, 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¦=60βπ₯ Let P = π₯π¦3 We need to maximize P Now, P = π₯π¦3 P = π₯(60βπ₯)3 Finding Pβ(x) P = π₯(60βπ₯)^3 Diff w.r.t π₯ ππ/ππ₯=π(π₯(60βπ₯)^3 )/ππ₯ ππ/ππ₯=π(π₯)/ππ₯ (60βπ₯)^3+(π(60βπ₯)^3)/ππ₯ . π₯ =(60βπ₯)^3+γ3(60βπ₯)γ^2 . (0β1) . π₯ =(60βπ₯)^3+3π₯(60βπ₯)^2 =(60βπ₯)^2 [(60βπ₯)β3π₯] =(60βπ₯)^2 [60β4π₯] Using product rule as (π’π£)^β²=π’^β² π£+π£^β² π’ Putting π π·/π π=π (60βπ₯)^2 (60β4π₯)=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 β15=45 Hence, numbers are 15 & 45