Ex 6.3,15 - Chapter 6 Class 12 Application of Derivatives

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Ex 6.3, 15 (Method 1) Find two positive numbers š‘„ and š‘¦ such that their sum is 35 and the product š‘„2 š‘¦5 is a maximum. Given two number are š‘„ & š‘¦ Such that š‘„ + š‘¦ = 35 š‘¦ = 35 ā€“ š‘„ Let P = š‘„2 š‘¦5 We need to maximize P Finding Pā€™(š’™) P(š‘„)=š‘„^2 š‘¦^5 P(š‘„)=š‘„^2 (35āˆ’š‘„)^5 Pā€™(š‘„)=š‘‘(š‘„^2 (35 āˆ’ š‘„)^5 )/š‘‘š‘„ Pā€™(š‘„)=š‘‘(š‘„^2 )/š‘‘š‘„ . (35āˆ’š‘„)^5+(š‘‘(35 āˆ’ š‘„)^5)/š‘‘š‘„ . š‘„^2 =2š‘„ .(35āˆ’š‘„)^5+怖5(35āˆ’š‘„)怗^4 .š‘‘(35 āˆ’ š‘„)/š‘‘š‘„ . š‘„^2 =2š‘„ .(35āˆ’š‘„)^5+怖5(35āˆ’š‘„)怗^4 . (0āˆ’1)(š‘„^2 ) =2š‘„ .(35āˆ’š‘„)^5+怖5(35āˆ’š‘„)怗^4 (āˆ’š‘„^2 ) =2š‘„ (35āˆ’š‘„)^5āˆ’怖5š‘„^2 (35āˆ’š‘„)怗^4 = 怖 š‘„ (35āˆ’š‘„)怗^4 [2(35āˆ’š‘„)āˆ’5š‘„] = 怖 š‘„ (35āˆ’š‘„)怗^4 (70āˆ’7š‘„) Putting Pā€™(š’™)=šŸŽ 怖 š‘„ (35āˆ’š‘„)怗^4 (70āˆ’7š‘„)=0 Hence š‘„ = 0 , 10 , 35 are Critical Points But, If we Take š‘„ = 0 Product will be 0 So, x = 0 is not possible If x = 35 š‘¦ = 35 ā€“ 35 = 35 ā€“ 35 = 0 So, product will be 0 So, x = 35 is not possible Hence only critical point is š‘„=10 Finding Pā€™ā€™(š’™) Pā€™(š‘„)=š‘„(35āˆ’š‘„)^4 (70āˆ’7š‘„) Pā€™(š‘„)=(35āˆ’š‘„)^4 (70š‘„āˆ’7š‘„^2 ) Pā€™ā€™(š‘„)=(š‘‘(35 āˆ’ š‘„)^4)/š‘‘š‘„. (70š‘„āˆ’7š‘„^2 )+š‘‘(70š‘„ āˆ’ 7š‘„^2 )/š‘‘š‘„ (35āˆ’š‘„)^4 =4(35āˆ’š‘„)^3.š‘‘(35 āˆ’ š‘„)/š‘‘š‘„. (70š‘„āˆ’7š‘„^2 )+(70āˆ’14š‘„) (35āˆ’š‘„)^4 =4(35āˆ’š‘„)^3 (0āˆ’1)(70š‘„āˆ’7š‘„^2 )+(70āˆ’14š‘„) (35āˆ’š‘„)^4 =āˆ’4(35āˆ’š‘„)^3 (70š‘„āˆ’7š‘„^2 )+(70āˆ’14š‘„) (35āˆ’š‘„)^4 Putting š‘„ = 10 in Pā€™ā€™(x) Pā€™ā€™(š‘„) = āˆ’4(35āˆ’š‘„)^3 (70š‘„āˆ’7š‘„^2 )+(70āˆ’14š‘„) (35āˆ’š‘„)^4 =āˆ’4(35āˆ’10)^3 (70(10)āˆ’7(10)^2 )+(70āˆ’14(10)) (35āˆ’10)^4 =āˆ’4(25)^3 (700āˆ’700)+(70āˆ’140) (25)^4 =āˆ’4(25)^3 (0)+(āˆ’70) (25)^4 =0āˆ’70(25)^4 =āˆ’70(25)^4 < 0 Thus, Pā€™ā€™(š‘„)<0 when š‘„ = 10 āˆ“ P is maximum when š‘„ = 10 Thus, when š‘„ = 10 š‘¦ = 35 ā€“ š‘„= 35 āˆ’10=25 Hence š’™ = 10 & š’š = 25 Ex 6.3, 15 (Method 2) Find two positive numbers š‘„ and š‘¦ such that their sum is 35 and the product š‘„2 š‘¦5 is a maximum. Given two number are š‘„ & š‘¦ Such that š‘„ + š‘¦ = 35 š‘¦ = 35 ā€“ š‘„ Let P = š‘„2 š‘¦5 We need to maximise P Finding Pā€™(š’™) P(š‘„)=š‘„^2 š‘¦^5 P(š‘„)=š‘„^2 (35āˆ’š‘„)^5 Pā€™(š‘„)=š‘‘(š‘„^2 (35 āˆ’ š‘„)^5 )/š‘‘š‘„ Pā€™(š‘„)=š‘‘(š‘„^2 )/š‘‘š‘„ . (35āˆ’š‘„)^5+(š‘‘(35 āˆ’ š‘„)^5)/š‘‘š‘„ . š‘„^2 =2š‘„ .(35āˆ’š‘„)^5+怖5(35āˆ’š‘„)怗^4 .š‘‘(35 āˆ’ š‘„)/š‘‘š‘„ . š‘„^2 =2š‘„ .(35āˆ’š‘„)^5+怖5(35āˆ’š‘„)怗^4 . (0āˆ’1)(š‘„^2 ) =2š‘„ .(35āˆ’š‘„)^5+怖5(35āˆ’š‘„)怗^4 (āˆ’š‘„^2 ) =2š‘„ (35āˆ’š‘„)^5āˆ’怖5š‘„^2 (35āˆ’š‘„)怗^4 = 怖 š‘„ (35āˆ’š‘„)怗^4 [2(35āˆ’š‘„)āˆ’5š‘„] = 怖 š‘„ (35āˆ’š‘„)怗^4 (70āˆ’7š‘„) Putting Pā€™(š’™)=šŸŽ 怖š‘„ (35āˆ’š‘„)怗^4 (70āˆ’7š‘„)=0 怖š‘„ (35āˆ’š‘„)怗^4 (70āˆ’7š‘„)=0 Hence š‘„ = 0 , 10 , 35 are Critical Points But, If We Take š‘„ = 0 Product will be 0 So, x = 0 is not possible If x = 35 š‘¦ = 35 ā€“ 35 = 35 ā€“ 35 = 0 So, product will be 0 So, x = 35 is not possible Hence only critical point is š‘„=10 āˆ“ š‘„ = 10 is point of maxima P(š‘„) is maximum at š‘„ = 10 Thus, when š‘„ = 10 š‘¦ = 35 ā€“ š‘„= 35 āˆ’10=25 Hence š’™ = 10 & š’š = 25

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo