Example 22 - Chapter 6 Class 12 Application of Derivatives
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Example 22 Find two positive numbers whose sum is 15 and the sum of whose squares is minimum. Let first number be 𝒙 Since Sum of two positive numbers is 15 𝑥+ 2nd number = 15 2nd number = 15 – 𝒙 Let S(𝑥) be the sum of the squares of the numbers S(𝑥)= (1st number)2 + (2nd number) 2 S(𝒙)=𝒙^𝟐+(𝟏𝟓−𝒙)^𝟐 We need to minimize S(𝒙) Finding S’(𝒙) S’(𝑥)=𝑑(𝑥^2+ (15 − 𝑥)^2 )/𝑑𝑥 =𝑑(𝑥^2 )/𝑑𝑥+(𝑑(15 − 𝑥)^2)/𝑑𝑥 = 2𝑥+ 2(15−𝑥)(−1) = 2𝑥− 2(15−𝑥) = 2𝑥−30+2𝑥 = 4𝒙−𝟑𝟎 Putting S’(𝒙)=𝟎 4𝑥−30=0 4𝑥=30 𝑥=30/4 𝒙=𝟏𝟓/𝟐 Finding S’’(𝒙) S’’(𝑥)=𝑑(4𝑥 − 30)/𝑑𝑥 = 4 Since S’’(𝒙)>𝟎 at 𝑥=15/2 ∴ 𝑥=15/2 is local minima Thus, S(𝑥) is Minimum at 𝑥=15/2 Hence, 1st number = 𝑥=𝟏𝟓/𝟐 2nd number = 15−𝑥=15−15/2=𝟏𝟓/𝟐
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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