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Ex 6.5, 16 - Find two positive numbers whose sum is 16 - Ex 6.5

Ex 6.5,16 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.5,16 - Chapter 6 Class 12 Application of Derivatives - Part 3
Ex 6.5,16 - Chapter 6 Class 12 Application of Derivatives - Part 4

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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 π‘₯ Now, First number + second number =16 π‘₯ + second number = 16 second number = 16 – π‘₯ Now, Sum of Cubes = (π‘“π‘–π‘Ÿπ‘ π‘‘ π‘›π‘’π‘šπ‘π‘’π‘Ÿ )^3+(π‘ π‘’π‘π‘œπ‘›π‘‘ π‘›π‘’π‘šπ‘π‘’π‘Ÿ )^3 Let S(π‘₯) = π‘₯3 + (16βˆ’π‘₯)^3 We Need to Find Minimum Value of s(π‘₯) Finding S’(π‘₯) S’(π‘₯)= 𝑑(π‘₯^3+ (16 βˆ’ π‘₯)^3 )/𝑑π‘₯ = 3π‘₯2 + 3(16βˆ’π‘₯)^2. (0βˆ’1) = 3π‘₯2 + 3(16βˆ’π‘₯)^2 (βˆ’1) = 3π‘₯2 – 3((16)^2+(π‘₯)^2βˆ’2(16)(π‘₯)) = 3π‘₯2 – 3(256+π‘₯^2βˆ’32π‘₯) = 3π‘₯2 – 3(256)βˆ’3π‘₯^2+3(32)π‘₯ = –3(256βˆ’32π‘₯) Putting S’(π‘₯)=0 –3(256βˆ’32π‘₯)=0 256 – 32π‘₯ = 0 32π‘₯ = 256 π‘₯ = 256/32 π‘₯ = 8 Finding S’’(π‘₯) S’(π‘₯)=βˆ’3(256βˆ’32π‘₯) S’’(π‘₯)=𝑑(βˆ’3(256 βˆ’ 32π‘₯))/𝑑π‘₯ = –3 𝑑(256 βˆ’ 32π‘₯)/𝑑π‘₯ = –3 [0βˆ’32] = 96 > 0 Since S’’(π‘₯)>0 for π‘₯ = 8 π‘₯ = 8 is point of local minima & S(π‘₯) is minimum at π‘₯ = 8 Hence, 1st number = x = 8 & 2nd number = 16 – x = 16 – 8 = 8

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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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.