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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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.