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Example 22 - Chapter 6 Class 12 Application of Derivatives
Last updated at Dec. 16, 2024 by Teachoo
Examples
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Transcript
Example 22 Find two positive numbers whose sum is 15 and the sum of whose squares is minimum. Let first number be š Since Sum of two positive numbers is 15 š„+ 2nd number = 15 2nd number = 15 ā š Let S(š„) be the sum of the squares of the numbers S(š„)= (1st number)2 + (2nd number) 2 S(š)=š^š+(ššāš)^š We need to minimize S(š) Finding Sā(š) Sā(š„)=š(š„^2+ (15 ā š„)^2 )/šš„ =š(š„^2 )/šš„+(š(15 ā š„)^2)/šš„ = 2š„+ 2(15āš„)(ā1) = 2š„ā 2(15āš„) = 2š„ā30+2š„ = 4šāšš Putting Sā(š)=š 4š„ā30=0 4š„=30 š„=30/4 š=šš/š Finding Sāā(š) Sāā(š„)=š(4š„ ā 30)/šš„ = 4 Since Sāā(š)>š at š„=15/2 ā“ š„=15/2 is local minima Thus, S(š„) is Minimum at š„=15/2 Hence, 1st number = š„=šš/š 2nd number = 15āš„=15ā15/2=šš/š
