Ex 6.3

Chapter 6 Class 12 Application of Derivatives
Serial order wise

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Ex 6.3, 15 (Method 1) Find two positive numbers π₯ and π¦ such that their sum is 35 and the product π₯2 π¦5 is a maximum. Given two number are π₯ & π¦ Such that π₯ + π¦ = 35 π¦ = 35 β π₯ Let P = π₯2 π¦5 We need to maximize P Finding Pβ(π) P(π₯)=π₯^2 π¦^5 P(π₯)=π₯^2 (35βπ₯)^5 Pβ(π₯)=π(π₯^2 (35 β π₯)^5 )/ππ₯ Pβ(π₯)=π(π₯^2 )/ππ₯ . (35βπ₯)^5+(π(35 β π₯)^5)/ππ₯ . π₯^2 =2π₯ .(35βπ₯)^5+γ5(35βπ₯)γ^4 .π(35 β π₯)/ππ₯ . π₯^2 =2π₯ .(35βπ₯)^5+γ5(35βπ₯)γ^4 . (0β1)(π₯^2 ) =2π₯ .(35βπ₯)^5+γ5(35βπ₯)γ^4 (βπ₯^2 ) =2π₯ (35βπ₯)^5βγ5π₯^2 (35βπ₯)γ^4 = γ π₯ (35βπ₯)γ^4 [2(35βπ₯)β5π₯] = γ π₯ (35βπ₯)γ^4 (70β7π₯) using product rule as (π’π£)^β²=π’^β² π£+π£^β² π’ Putting Pβ(π)=π γ π₯ (35βπ₯)γ^4 (70β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 (35βπ₯)^4= 0 35βπ₯=0 π₯=35 70β7π₯= 0 7π₯=70 π₯=70/7 π₯= 10 If x = 35 π¦ = 35 β 35 = 35 β 35 = 0 So, product will be 0 So, x = 35 is not possible Hence only critical point is π₯=10 Finding Pββ(π) Pβ(π₯)=π₯(35βπ₯)^4 (70β7π₯) Pβ(π₯)=(35βπ₯)^4 (70π₯β7π₯^2 ) Pββ(π₯)=(π(35 β π₯)^4)/ππ₯. (70π₯β7π₯^2 )+π(70π₯ β 7π₯^2 )/ππ₯ (35βπ₯)^4 =4(35βπ₯)^3.π(35 β π₯)/ππ₯. (70π₯β7π₯^2 )+(70β14π₯) (35βπ₯)^4 Using product rule as (π’π£)^β²=π’^β² π£+π£^β² π’ =4(35βπ₯)^3 (0β1)(70π₯β7π₯^2 )+(70β14π₯) (35βπ₯)^4 =β4(35βπ₯)^3 (70π₯β7π₯^2 )+(70β14π₯) (35βπ₯)^4 Putting π₯ = 10 in Pββ(x) Pββ(π₯) = β4(35βπ₯)^3 (70π₯β7π₯^2 )+(70β14π₯) (35βπ₯)^4 =β4(35β10)^3 (70(10)β7(10)^2 )+(70β14(10)) (35β10)^4 =β4(25)^3 (700β700)+(70β140) (25)^4 =β4(25)^3 (0)+(β70) (25)^4 =0β70(25)^4 =β70(25)^4 < 0 Thus, Pββ(π₯)<0 when π₯ = 10 β΄ P is maximum when π₯ = 10 Thus, when π₯ = 10 π¦ = 35 β π₯= 35 β10=25 Hence π = 10 & π = 25 Ex 6.3, 15 (Method 2) Find two positive numbers π₯ and π¦ such that their sum is 35 and the product π₯2 π¦5 is a maximum. Given two number are π₯ & π¦ Such that π₯ + π¦ = 35 π¦ = 35 β π₯ Let P = π₯2 π¦5 We need to maximise P Finding Pβ(π) P(π₯)=π₯^2 π¦^5 P(π₯)=π₯^2 (35βπ₯)^5 Pβ(π₯)=π(π₯^2 (35 β π₯)^5 )/ππ₯ Pβ(π₯)=π(π₯^2 )/ππ₯ . (35βπ₯)^5+(π(35 β π₯)^5)/ππ₯ . π₯^2 =2π₯ .(35βπ₯)^5+γ5(35βπ₯)γ^4 .π(35 β π₯)/ππ₯ . π₯^2 =2π₯ .(35βπ₯)^5+γ5(35βπ₯)γ^4 . (0β1)(π₯^2 ) =2π₯ .(35βπ₯)^5+γ5(35βπ₯)γ^4 (βπ₯^2 ) =2π₯ (35βπ₯)^5βγ5π₯^2 (35βπ₯)γ^4 = γ π₯ (35βπ₯)γ^4 [2(35βπ₯)β5π₯] = γ π₯ (35βπ₯)γ^4 (70β7π₯) Using product rule as (π’π£)^β²=π’^β² π£+π£^β² π’ Putting Pβ(π)=π γπ₯ (35βπ₯)γ^4 (70β7π₯)=0 γπ₯ (35βπ₯)γ^4 (70β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 (35βπ₯)^4= 0 35βπ₯=0 π₯=35 70β7π₯= 0 7π₯=70 π₯=70/7 π₯= 10 If x = 35 π¦ = 35 β 35 = 35 β 35 = 0 So, product will be 0 So, x = 35 is not possible Hence only critical point is π₯=10 β΄ π₯ = 10 is point of maxima P(π₯) is maximum at π₯ = 10 Thus, when π₯ = 10 π¦ = 35 β π₯= 35 β10=25 Hence π = 10 & π = 25