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Ex 6.5, 13 - Find two numbers whose sum is 24, product is large

Ex 6.5,13 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.5,13 - Chapter 6 Class 12 Application of Derivatives - Part 3 Ex 6.5,13 - Chapter 6 Class 12 Application of Derivatives - Part 4

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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 – π‘₯ Product = (π‘“π‘–π‘Ÿπ‘ π‘‘ π‘›π‘’π‘šπ‘π‘’π‘Ÿ ) Γ— (π‘ π‘’π‘π‘œπ‘›π‘‘ π‘›π‘’π‘šπ‘π‘’π‘Ÿ) = π‘₯ (24βˆ’π‘₯) Let P(π‘₯) = π‘₯ (24βˆ’π‘₯) We need product as large as possible Hence we need to find maximum value of P(π‘₯) Finding P’(x) P(π‘₯)=π‘₯(24βˆ’π‘₯) P(π‘₯)=24π‘₯βˆ’π‘₯^2 P’(π‘₯)=24βˆ’2π‘₯ P’(π‘₯)=2(12βˆ’π‘₯) Putting P’(π‘₯)=0 2(12βˆ’π‘₯)=0 12 – π‘₯ = 0 π‘₯ = 12 Finding P’’(π‘₯) P’(π‘₯)=24βˆ’2π‘₯ P’’(π‘₯) = 0 – 2 = – 2 Thus, p’’(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‘₯(24βˆ’π‘₯) Putting π‘₯ = 12 p(12)= 12(24βˆ’12) = 12(12) = 144 ∴ First number = x = 12 & Second number = 24 – x = (24 – 12)= 12

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.