Example 22 - Chapter 6 Class 12 Application of Derivatives
Last updated at August 17, 2026 by Teachoo
Examples
Example 1
You are here
You are here
Chapter 6 Class 12 Application of Derivatives
Serial order wise
Transcript
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
