Ex 6.5,16 - Chapter 6 Class 12 Application of Derivatives

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