Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

Ex 6.3

Ex 6.3, 1 (i)
Important

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 June 12, 2023 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 𝑥 Now, First number + second number =16 𝑥 + second number = 16 second number = 16 – 𝑥 Now, Sum of Cubes = (𝑓𝑖𝑟𝑠𝑡 𝑛𝑢𝑚𝑏𝑒𝑟 )^3+(𝑠𝑒𝑐𝑜𝑛𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 )^3 Let S(𝑥) = 𝑥3 + (16−𝑥)^3 We Need to Find Minimum Value of s(𝑥) Finding S’(𝑥) S’(𝑥)= 𝑑(𝑥^3+ (16 − 𝑥)^3 )/𝑑𝑥 = 3𝑥2 + 3(16−𝑥)^2. (0−1) = 3𝑥2 + 3(16−𝑥)^2 (−1) = 3𝑥2 – 3((16)^2+(𝑥)^2−2(16)(𝑥)) = 3𝑥2 – 3(256+𝑥^2−32𝑥) = 3𝑥2 – 3(256)−3𝑥^2+3(32)𝑥 = –3(256−32𝑥) Putting S’(𝑥)=0 –3(256−32𝑥)=0 256 – 32𝑥 = 0 32𝑥 = 256 𝑥 = 256/32 𝑥 = 8 Finding S’’(𝑥) S’(𝑥)=−3(256−32𝑥) S’’(𝑥)=𝑑(−3(256 − 32𝑥))/𝑑𝑥 = –3 𝑑(256 − 32𝑥)/𝑑𝑥 = –3 [0−32] = 96 > 0 Since S’’(𝑥)>0 for 𝑥 = 8 𝑥 = 8 is point of local minima & S(𝑥) is minimum at 𝑥 = 8 Hence, 1st number = x = 8 & 2nd number = 16 – x = 16 – 8 = 8