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

Last updated at Dec. 16, 2024 by Teachoo

 

Transcript

Ex 6.3, 13 Find two numbers whose sum is 24 and whose product is as large as possible. Let first number be 𝑥 Now, given that First number + Second number = 24 𝑥 + second number = 24 Second number = 24 – 𝑥 Product = (𝑓𝑖𝑟𝑠𝑡 𝑛𝑢𝑚𝑏𝑒𝑟 ) × (𝑠𝑒𝑐𝑜𝑛𝑑 𝑛𝑢𝑚𝑏𝑒𝑟) = 𝑥 (24−𝑥) Let P(𝑥) = 𝑥 (24−𝑥) We need product as large as possible Hence we need to find maximum value of P(𝑥) Finding P’(x) P(𝑥)=𝑥(24−𝑥) P(𝑥)=24𝑥−𝑥^2 P’(𝑥)=24−2𝑥 P’(𝑥)=2(12−𝑥) Putting P’(𝑥)=0 2(12−𝑥)=0 12 – 𝑥 = 0 𝑥 = 12 Finding P’’(𝑥) P’(𝑥)=24−2𝑥 P’’(𝑥) = 0 – 2 = – 2 Thus, p’’(𝑥) < 0 for 𝑥 = 12 𝑥 = 12 is point of maxima & P(𝑥) is maximum at 𝑥 = 12 Finding maximum P(x) P(𝑥)=𝑥(24−𝑥) Putting 𝑥 = 12 p(12)= 12(24−12) = 12(12) = 144 ∴ First number = x = 12 & Second number = 24 – x = (24 – 12)= 12
  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise

    3. Ex 6.3

    Examples →
Examples →
Chapter 6 Class 12 Application of Derivatives
Serial order wise

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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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo