Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12

Last updated at May 29, 2018 by Teachoo

Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12

Transcript

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

Ex 6.5

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Ex 6.5,16 You are here

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Chapter 6 Class 12 Application of Derivatives

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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.