Ex 6.5, 14 - Find x and y such that x + y = 60, xy3 is max - Minima/ maxima (statement questions) - Number questions

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  1. Chapter 6 Class 12 Application of Derivatives
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Ex 6.5,14 (Method 1) Find two positive numbers š‘„ and y such that š‘„ + š‘¦ = 60 and š‘„š‘¦3 is maximum. Given two number š‘„ and y, such that š‘„ + š‘¦ = 60 š‘¦=60āˆ’š‘„ Let P = š‘„š‘¦3 We need to maximize P Now, P = š‘„š‘¦3 P = š‘„ļ·60āˆ’š‘„ļ·Æ3 Step 1: Finding Pā€™(x) P = š‘„ļ·ļ·60āˆ’š‘„ļ·Æļ·®3ļ·Æ Diff w.r.t š‘„ ļ·š‘‘š‘ļ·®š‘‘š‘„ļ·Æ=ļ·š‘‘ļ·š‘„ļ·ļ·60āˆ’š‘„ļ·Æļ·®3ļ·Æļ·Æļ·®š‘‘š‘„ļ·Æ ļ·š‘‘š‘ļ·®š‘‘š‘„ļ·Æ=ļ·š‘‘ļ·š‘„ļ·Æļ·®š‘‘š‘„ļ·Æļ·ļ·60āˆ’š‘„ļ·Æļ·®3ļ·Æ+ļ·š‘‘ļ·ļ·60āˆ’š‘„ļ·Æļ·®3ļ·Æļ·®š‘‘š‘„ļ·Æ . š‘„ =ļ·ļ·60āˆ’š‘„ļ·Æļ·®3ļ·Æ+ļ·3ļ·60āˆ’š‘„ļ·Æļ·®2ļ·Æ . ļ·0āˆ’1ļ·Æ . š‘„ =ļ·ļ·60āˆ’š‘„ļ·Æļ·®3ļ·Æ+3š‘„ļ·ļ·60āˆ’š‘„ļ·Æļ·®2ļ·Æ =ļ·ļ·60āˆ’š‘„ļ·Æļ·®2ļ·Æļ·ļ·60āˆ’š‘„ļ·Æāˆ’3š‘„ļ·Æ =ļ·ļ·60āˆ’š‘„ļ·Æļ·®2ļ·Æļ·60āˆ’4š‘„ļ·Æ Step 2: Putting ļ·š‘‘š‘ƒļ·®š‘‘š‘„ļ·Æ=0 ļ·ļ·60āˆ’š‘„ļ·Æļ·®2ļ·Æļ·60āˆ’4š‘„ļ·Æ=0 So, x = 60 & x = ļ·60ļ·®4ļ·Æ = 15 But, If š‘„=60 š‘¦= 60 ā€“ 60 = 0 Which is not possible Hence š‘„= 15 is only critical point. Step 3: Finding Pā€™ā€™ ļ·š‘„ļ·Æ Pā€™ā€™ ļ·š‘„ļ·Æ=ļ·š‘‘ļ·ļ·ļ·60 āˆ’ š‘„ļ·Æļ·®2ļ·Æ60 āˆ’ 4š‘„ļ·Æļ·®š‘‘š‘„ļ·Æ Pā€™ā€™ ļ·š‘„ļ·Æ=ļ·š‘‘ļ·ļ·60 āˆ’ š‘„ļ·Æļ·®2ļ·Æļ·®š‘‘š‘„ļ·Æ . ļ·60āˆ’4š‘„ļ·Æ+ļ·š‘‘ļ·60 āˆ’ 4š‘„ļ·Æļ·®š‘‘š‘„ļ·Æļ·ļ·60āˆ’š‘„ļ·Æļ·®2ļ·Æ = 2ļ·60āˆ’š‘„ļ·Æ .ļ·0āˆ’1ļ·Æļ·60āˆ’4š‘„ļ·Æāˆ’4ļ·ļ·60āˆ’š‘„ļ·Æļ·®2ļ·Æ = āˆ’2ļ·60āˆ’š‘„ļ·Æ . ļ·60āˆ’4š‘„ļ·Æāˆ’4ļ·ļ·60āˆ’š‘„ļ·Æļ·®2ļ·Æ = āˆ’2ļ·60āˆ’š‘„ļ·Æļ·ļ·60āˆ’4š‘„ļ·Æ+2ļ·60āˆ’š‘„ļ·Æļ·Æ = āˆ’2ļ·60āˆ’š‘„ļ·Æļ·ļ·60āˆ’4š‘„ļ·Æ+120āˆ’2š‘„ļ·Æ = āˆ’2ļ·60āˆ’š‘„ļ·Æļ·180āˆ’6š‘„ļ·Æ At š‘„ = 15 Pā€™ā€™ļ·15ļ·Æ=āˆ’2ļ·60āˆ’15ļ·Æļ·180āˆ’6ļ·15ļ·Æļ·Æ=āˆ’90 Ɨ90=āˆ’8100 < 0 āˆ“ Pā€™ā€™ļ·š‘„ļ·Æ<0 at š‘„ = 15 Hence š‘„š‘¦3 is Maximum when š‘„ = 15 Thus, when š‘„ = 15 š‘¦ =60 ā€“ š‘„=60 āˆ’15=45 Hence, numbers are 15 & 45 Ex 6.5,14 (Method 2) Find two positive numbers š‘„ and y such that š‘„ + š‘¦ = 60 and š‘„š‘¦3 is maximum. Given two number š‘„ and y, such that š‘„ + š‘¦ = 60 š‘¦=60āˆ’š‘„ Let P = š‘„š‘¦3 We need to maximize P Now, P = š‘„š‘¦3 P = š‘„ļ·60āˆ’š‘„ļ·Æ3 Step 1: Finding Pā€™(x) P = š‘„ļ·ļ·60āˆ’š‘„ļ·Æļ·®3ļ·Æ Diff w.r.t š‘„ ļ·š‘‘š‘ļ·®š‘‘š‘„ļ·Æ=ļ·š‘‘ļ·š‘„ļ·ļ·60āˆ’š‘„ļ·Æļ·®3ļ·Æļ·Æļ·®š‘‘š‘„ļ·Æ ļ·š‘‘š‘ļ·®š‘‘š‘„ļ·Æ=ļ·š‘‘ļ·š‘„ļ·Æļ·®š‘‘š‘„ļ·Æļ·ļ·60āˆ’š‘„ļ·Æļ·®3ļ·Æ+ļ·š‘‘ļ·ļ·60āˆ’š‘„ļ·Æļ·®3ļ·Æļ·®š‘‘š‘„ļ·Æ . š‘„ =ļ·ļ·60āˆ’š‘„ļ·Æļ·®3ļ·Æ+ļ·3ļ·60āˆ’š‘„ļ·Æļ·®2ļ·Æ . ļ·0āˆ’1ļ·Æ . š‘„ =ļ·ļ·60āˆ’š‘„ļ·Æļ·®3ļ·Æ+3š‘„ļ·ļ·60āˆ’š‘„ļ·Æļ·®2ļ·Æ =ļ·ļ·60āˆ’š‘„ļ·Æļ·®2ļ·Æļ·ļ·60āˆ’š‘„ļ·Æāˆ’3š‘„ļ·Æ =ļ·ļ·60āˆ’š‘„ļ·Æļ·®2ļ·Æļ·60āˆ’4š‘„ļ·Æ Step 2: Putting ļ·š‘‘š‘ƒļ·®š‘‘š‘„ļ·Æ=0 ļ·ļ·60āˆ’š‘„ļ·Æļ·®2ļ·Æļ·60āˆ’4š‘„ļ·Æ=0 So, x = 60 & x = ļ·60ļ·®4ļ·Æ = 15 But, If š‘„=60 š‘¦= 60 ā€“ 60 = 0 Which is not possible Hence š‘„= 15 is only critical point. Step 3: ā‡’ š‘„ = 15 is point of local Maxima & Pļ·š‘„ļ·Æ is Maximum at š‘„ = 15 Thus, when š‘„ = 15 š‘¦ =60 ā€“ š‘„=60 āˆ’15=45 Hence, numbers are 15 & 45

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