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
Ex 6.5,2 Important
Ex 6.5,3
Ex 6.5,4
Ex 6.5,5 Important
Ex 6.5,6
Ex 6.5,7 Important
Ex 6.5,8
Ex 6.5,9 Important
Ex 6.5,10
Ex 6.5,11 Important
Ex 6.5,12 Important
Ex 6.5,13
Ex 6.5,14 Important
Ex 6.5,15 Important
Ex 6.5,16 You are here
Ex 6.5,17
Ex 6.5,18 Important
Ex 6.5,19 Important
Ex 6.5, 20 Important
Ex 6.5,21
Ex 6.5,22 Important
Ex 6.5,23 Important
Ex 6.5,24 Important
Ex 6.5,25 Important
Ex 6.5,26 Important
Ex 6.5, 27
Ex 6.5,28 Important
Ex 6.5,29
About the Author