Ex 6.5,13 - Chapter 6 Class 12 Application of Derivatives (Term 1)
Last updated at April 15, 2021 by Teachoo
Minima/ maxima (statement questions) - Number questions
Minima/ maxima (statement questions) - Number questions
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