Ex 6.3,16 - Chapter 6 Class 12 Application of Derivatives
Last updated at April 16, 2024 by Teachoo
Ex 6.3
Ex 6.3, 1 (ii)
Ex 6.3, 1 (iii) Important
Ex 6.3, 1 (iv)
Ex 6.3, 2 (i)
Ex 6.3, 2 (ii) Important
Ex 6.3, 2 (iii)
Ex 6.3, 2 (iv) Important
Ex 6.3, 2 (v) Important
Ex 6.3, 3 (i)
Ex 6.3, 3 (ii)
Ex 6.3, 3 (iii)
Ex 6.3, 3 (iv) Important
Ex 6.3, 3 (v)
Ex 6.3, 3 (vi)
Ex 6.3, 3 (vii) Important
Ex 6.3, 3 (viii)
Ex 6.3, 4 (i)
Ex 6.3, 4 (ii) Important
Ex 6.3, 4 (iii)
Ex 6.3, 5 (i)
Ex 6.3, 5 (ii)
Ex 6.3, 5 (iii) Important
Ex 6.3, 5 (iv)
Ex 6.3,6
Ex 6.3,7 Important
Ex 6.3,8
Ex 6.3,9 Important
Ex 6.3,10
Ex 6.3,11 Important
Ex 6.3,12 Important
Ex 6.3,13
Ex 6.3,14 Important
Ex 6.3,15 Important
Ex 6.3,16 You are here
Ex 6.3,17
Ex 6.3,18 Important
Ex 6.3,19 Important
Ex 6.3, 20 Important
Ex 6.3,21
Ex 6.3,22 Important
Ex 6.3,23 Important
Ex 6.3,24 Important
Ex 6.3,25 Important
Ex 6.3, 26 Important
Ex 6.3, 27 (MCQ)
Ex 6.3,28 (MCQ) Important
Ex 6.3,29 (MCQ)
Last updated at April 16, 2024 by Teachoo
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 = (šššš š” šš¢šššš )^3+(š ššššš šš¢šššš )^3 Let S(š„) = š„3 + (16āš„)^3 We Need to Find Minimum Value of s(š„) Finding Sā(š„) Sā(š„)= š(š„^3+ (16 ā š„)^3 )/šš„ = 3š„2 + 3(16āš„)^2. (0ā1) = 3š„2 + 3(16āš„)^2 (ā1) = 3š„2 ā 3((16)^2+(š„)^2ā2(16)(š„)) = 3š„2 ā 3(256+š„^2ā32š„) = 3š„2 ā 3(256)ā3š„^2+3(32)š„ = ā3(256ā32š„) Putting Sā(š„)=0 ā3(256ā32š„)=0 256 ā 32š„ = 0 32š„ = 256 š„ = 256/32 š„ = 8 Finding Sāā(š„) Sā(š„)=ā3(256ā32š„) Sāā(š„)=š(ā3(256 ā 32š„))/šš„ = ā3 š(256 ā 32š„)/šš„ = ā3 [0ā32] = 96 > 0 Since Sāā(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