# Ex 6.5,14

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

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¦=60āš„ Let P = š„š¦3 We need to maximize P Now, P = š„š¦3 P = š„ļ·60āš„ļ·Æ3 Step 1: Finding Pā(x) P = š„ļ·ļ·60āš„ļ·Æļ·®3ļ·Æ Diff w.r.t š„ ļ·ššļ·®šš„ļ·Æ=ļ·šļ·š„ļ·ļ·60āš„ļ·Æļ·®3ļ·Æļ·Æļ·®šš„ļ·Æ ļ·ššļ·®šš„ļ·Æ=ļ·šļ·š„ļ·Æļ·®šš„ļ·Æļ·ļ·60āš„ļ·Æļ·®3ļ·Æ+ļ·šļ·ļ·60āš„ļ·Æļ·®3ļ·Æļ·®šš„ļ·Æ . š„ =ļ·ļ·60āš„ļ·Æļ·®3ļ·Æ+ļ·3ļ·60āš„ļ·Æļ·®2ļ·Æ . ļ·0ā1ļ·Æ . š„ =ļ·ļ·60āš„ļ·Æļ·®3ļ·Æ+3š„ļ·ļ·60āš„ļ·Æļ·®2ļ·Æ =ļ·ļ·60āš„ļ·Æļ·®2ļ·Æļ·ļ·60āš„ļ·Æā3š„ļ·Æ =ļ·ļ·60āš„ļ·Æļ·®2ļ·Æļ·60ā4š„ļ·Æ Step 2: Putting ļ·ššļ·®šš„ļ·Æ=0 ļ·ļ·60āš„ļ·Æļ·®2ļ·Æļ·60ā4š„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 Pāā ļ·š„ļ·Æ Pāā ļ·š„ļ·Æ=ļ·šļ·ļ·ļ·60 ā š„ļ·Æļ·®2ļ·Æ60 ā 4š„ļ·Æļ·®šš„ļ·Æ Pāā ļ·š„ļ·Æ=ļ·šļ·ļ·60 ā š„ļ·Æļ·®2ļ·Æļ·®šš„ļ·Æ . ļ·60ā4š„ļ·Æ+ļ·šļ·60 ā 4š„ļ·Æļ·®šš„ļ·Æļ·ļ·60āš„ļ·Æļ·®2ļ·Æ = 2ļ·60āš„ļ·Æ .ļ·0ā1ļ·Æļ·60ā4š„ļ·Æā4ļ·ļ·60āš„ļ·Æļ·®2ļ·Æ = ā2ļ·60āš„ļ·Æ . ļ·60ā4š„ļ·Æā4ļ·ļ·60āš„ļ·Æļ·®2ļ·Æ = ā2ļ·60āš„ļ·Æļ·ļ·60ā4š„ļ·Æ+2ļ·60āš„ļ·Æļ·Æ = ā2ļ·60āš„ļ·Æļ·ļ·60ā4š„ļ·Æ+120ā2š„ļ·Æ = ā2ļ·60āš„ļ·Æļ·180ā6š„ļ·Æ At š„ = 15 Pāāļ·15ļ·Æ=ā2ļ·60ā15ļ·Æļ·180ā6ļ·15ļ·Æļ·Æ=ā90 Ć90=ā8100 < 0 ā“ Pāāļ·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 ā15=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¦=60āš„ Let P = š„š¦3 We need to maximize P Now, P = š„š¦3 P = š„ļ·60āš„ļ·Æ3 Step 1: Finding Pā(x) P = š„ļ·ļ·60āš„ļ·Æļ·®3ļ·Æ Diff w.r.t š„ ļ·ššļ·®šš„ļ·Æ=ļ·šļ·š„ļ·ļ·60āš„ļ·Æļ·®3ļ·Æļ·Æļ·®šš„ļ·Æ ļ·ššļ·®šš„ļ·Æ=ļ·šļ·š„ļ·Æļ·®šš„ļ·Æļ·ļ·60āš„ļ·Æļ·®3ļ·Æ+ļ·šļ·ļ·60āš„ļ·Æļ·®3ļ·Æļ·®šš„ļ·Æ . š„ =ļ·ļ·60āš„ļ·Æļ·®3ļ·Æ+ļ·3ļ·60āš„ļ·Æļ·®2ļ·Æ . ļ·0ā1ļ·Æ . š„ =ļ·ļ·60āš„ļ·Æļ·®3ļ·Æ+3š„ļ·ļ·60āš„ļ·Æļ·®2ļ·Æ =ļ·ļ·60āš„ļ·Æļ·®2ļ·Æļ·ļ·60āš„ļ·Æā3š„ļ·Æ =ļ·ļ·60āš„ļ·Æļ·®2ļ·Æļ·60ā4š„ļ·Æ Step 2: Putting ļ·ššļ·®šš„ļ·Æ=0 ļ·ļ·60āš„ļ·Æļ·®2ļ·Æļ·60ā4š„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 ā15=45 Hence, numbers are 15 & 45

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Ex 6.5,14 You are here

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Chapter 6 Class 12 Application of Derivatives

Serial order wise

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