# Ex 6.5,13 - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at April 15, 2021 by Teachoo

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Ex 6.5,13 You are here

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Ex 6.5, 27 (MCQ)

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Last updated at April 15, 2021 by Teachoo

Ex 6.5, 13 Find two numbers whose sum is 24 and whose product is as large as possible. Let first number be π₯ Now, given that First number + Second number = 24 π₯ + second number = 24 Second number = 24 β π₯ Product = (ππππ π‘ ππ’ππππ ) Γ (π πππππ ππ’ππππ) = π₯ (24βπ₯) Let P(π₯) = π₯ (24βπ₯) We need product as large as possible Hence we need to find maximum value of P(π₯) Finding Pβ(x) P(π₯)=π₯(24βπ₯) P(π₯)=24π₯βπ₯^2 Pβ(π₯)=24β2π₯ Pβ(π₯)=2(12βπ₯) Putting Pβ(π₯)=0 2(12βπ₯)=0 12 β π₯ = 0 π₯ = 12 Finding Pββ(π₯) Pβ(π₯)=24β2π₯ Pββ(π₯) = 0 β 2 = β 2 Thus, pββ(π₯) < 0 for π₯ = 12 π₯ = 12 is point of maxima & P(π₯) is maximum at π₯ = 12 Finding maximum P(x) P(π₯)=π₯(24βπ₯) Putting π₯ = 12 p(12)= 12(24β12) = 12(12) = 144 β΄ First number = x = 12 & Second number = 24 β x = (24 β 12)= 12