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Last updated at May 29, 2018 by Teachoo

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Ex 6.5,16 Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum. Let first number be š„ Now, First number + second number =16 š„ + second number = 16 second number = 16 ā š„ Now, Sum of Cubes = ļ·ļ·šššš š” šš¢šššš ļ·Æļ·®3ļ·Æ+ļ·ļ·š ššššš šš¢šššš ļ·Æļ·®3ļ·Æ Let Sļ·š„ļ·Æ = š„3 + ļ·ļ·16āš„ļ·Æļ·®3ļ·Æ We Need to Find Minimum Value of sļ·š„ļ·Æ Finding Sāļ·š„ļ·Æ Sāļ·š„ļ·Æ= ļ·šļ·ļ·š„ļ·®3ļ·Æ+ ļ·ļ·16 ā š„ļ·Æļ·®3ļ·Æļ·Æļ·®šš„ļ·Æ = 3š„2 + 3ļ·ļ·16āš„ļ·Æļ·®2ļ·Æ. ļ·0ā1ļ·Æ = 3š„2 + 3ļ·ļ·16āš„ļ·Æļ·®2ļ·Æļ·ā1ļ·Æ = 3š„2 ā 3ļ·ļ·ļ·16ļ·Æļ·®2ļ·Æ+ļ·ļ·š„ļ·Æļ·®2ļ·Æā2ļ·16ļ·Æļ·š„ļ·Æļ·Æ = 3š„2 ā 3ļ·256+ļ·š„ļ·®2ļ·Æā32š„ļ·Æ = 3š„2 ā 3ļ·256ļ·Æā3ļ·š„ļ·®2ļ·Æ+3ļ·32ļ·Æš„ = ā3ļ·256ā32š„ļ·Æ Putting Sāļ·š„ļ·Æ=0 ā3ļ·256ā32š„ļ·Æ=0 256 ā 32š„ = 0 32š„ = 256 š„ = ļ·256ļ·®32ļ·Æ š„ = 8 Finding Sāāļ·š„ļ·Æ Sāļ·š„ļ·Æ=ā3ļ·256ā32š„ļ·Æ Sāāļ·š„ļ·Æ=ļ·šļ·ā3ļ·256 ā 32š„ļ·Æļ·Æļ·®šš„ļ·Æ = ā3 ļ·šļ·256 ā 32š„ļ·Æļ·®šš„ļ·Æ = ā3 ļ·0ā32ļ·Æ = 96 > 0 Since Sāāļ·š„ļ·Æ>0 for š„ = 8 ā š„ = 8 is point of local minima & Sļ·š„ļ·Æ is minimum at š„ = 8 Hence, 1st number = x = 8 & 2nd number = 16 ā x = 16 ā 8 = 8

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Chapter 6 Class 12 Application of Derivatives

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.