Ex 6.5

Chapter 6 Class 12 Application of Derivatives
Serial order wise

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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 β π₯ Now, Sum of Cubes = (ππππ π‘ ππ’ππππ )^3+(π πππππ ππ’ππππ )^3 Let S(π₯) = π₯3 + (16βπ₯)^3 We Need to Find Minimum Value of s(π₯) Finding Sβ(π₯) Sβ(π₯)= π(π₯^3+ (16 β π₯)^3 )/ππ₯ = 3π₯2 + 3(16βπ₯)^2. (0β1) = 3π₯2 + 3(16βπ₯)^2 (β1) = 3π₯2 β 3((16)^2+(π₯)^2β2(16)(π₯)) = 3π₯2 β 3(256+π₯^2β32π₯) = 3π₯2 β 3(256)β3π₯^2+3(32)π₯ = β3(256β32π₯) Putting Sβ(π₯)=0 β3(256β32π₯)=0 256 β 32π₯ = 0 32π₯ = 256 π₯ = 256/32 π₯ = 8 Finding Sββ(π₯) Sβ(π₯)=β3(256β32π₯) Sββ(π₯)=π(β3(256 β 32π₯))/ππ₯ = β3 π(256 β 32π₯)/ππ₯ = β3 [0β32] = 96 > 0 Since Sββ(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 β x = 16 β 8 = 8