Ex 6.3,13 - Chapter 6 Class 12 Application of Derivatives

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Ex 6.3, 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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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