Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12    1. Chapter 6 Class 12 Application of Derivatives
2. Serial order wise
3. Ex 6.5

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 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﷯−3﷐𝑥﷮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

Ex 6.5 