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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


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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.