Ex 6.3

Chapter 6 Class 12 Application of Derivatives
Serial order wise

### Transcript

Ex 6.3, 16 Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.Let first number be š„ Now, First number + second number =16 š„ + second number = 16 second number = 16 ā š„ Now, Sum of Cubes = (šššš š” šš¢šššš )^3+(š ššššš šš¢šššš )^3 Let S(š„) = š„3 + (16āš„)^3 We Need to Find Minimum Value of s(š„) Finding Sā(š„) Sā(š„)= š(š„^3+ (16 ā š„)^3 )/šš„ = 3š„2 + 3(16āš„)^2. (0ā1) = 3š„2 + 3(16āš„)^2 (ā1) = 3š„2 ā 3((16)^2+(š„)^2ā2(16)(š„)) = 3š„2 ā 3(256+š„^2ā32š„) = 3š„2 ā 3(256)ā3š„^2+3(32)š„ = ā3(256ā32š„) Putting Sā(š„)=0 ā3(256ā32š„)=0 256 ā 32š„ = 0 32š„ = 256 š„ = 256/32 š„ = 8 Finding Sāā(š„) Sā(š„)=ā3(256ā32š„) Sāā(š„)=š(ā3(256 ā 32š„))/šš„ = ā3 š(256 ā 32š„)/šš„ = ā3 [0ā32] = 96 > 0 Since Sāā(š„)>0 for š„ = 8 š„ = 8 is point of local minima & S(š„) is minimum at š„ = 8 Hence, 1st number = x = 8 & 2nd number = 16 ā x = 16 ā 8 = 8