Ex 6.5, 15 - Find x and y, sum is 35 and x2 y5 is maximum - Minima/ maxima (statement questions) - Number questions

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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise
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Ex 6.5,15 (Method 1) Find two positive numbers 𝑥 and 𝑦 such that their sum is 35 and the product 𝑥2 𝑦5 is a maximum. Given two number are 𝑥 & 𝑦 Such that 𝑥 + 𝑦 = 35 𝑦 = 35 – 𝑥 Let P = 𝑥2 𝑦5 We need to maximise P Step 1: Finding P’﷐𝑥﷯ P﷐𝑥﷯=﷐𝑥﷮2﷯﷐𝑦﷮5﷯ P﷐𝑥﷯=﷐𝑥﷮2﷯﷐﷐35−𝑥﷯﷮5﷯ P’﷐𝑥﷯=﷐𝑑﷐﷐𝑥﷮2﷯﷐﷐35 − 𝑥﷯﷮5﷯﷯﷮𝑑𝑥﷯ P’﷐𝑥﷯=﷐𝑑﷐﷐𝑥﷮2﷯﷯﷮𝑑𝑥﷯ . ﷐﷐35−𝑥﷯﷮5﷯+﷐𝑑﷐﷐35 − 𝑥﷯﷮5﷯﷮𝑑𝑥﷯ . ﷐𝑥﷮2﷯ =2𝑥 .﷐﷐35−𝑥﷯﷮5﷯+﷐5﷐35−𝑥﷯﷮4﷯ .﷐𝑑﷐35 − 𝑥﷯﷮𝑑𝑥﷯ . ﷐𝑥﷮2﷯ =2𝑥 .﷐﷐35−𝑥﷯﷮5﷯+﷐5﷐35−𝑥﷯﷮4﷯ . ﷐0−1﷯﷐﷐𝑥﷮2﷯﷯ =2𝑥 .﷐﷐35−𝑥﷯﷮5﷯+﷐5﷐35−𝑥﷯﷮4﷯ ﷐−﷐𝑥﷮2﷯﷯ =2𝑥 ﷐﷐35−𝑥﷯﷮5﷯−﷐5﷐𝑥﷮2﷯ ﷐35−𝑥﷯﷮4﷯ = ﷐ 𝑥 ﷐35−𝑥﷯﷮4﷯﷐2﷐35−𝑥﷯−5𝑥﷯ = ﷐ 𝑥 ﷐35−𝑥﷯﷮4﷯﷐70−7𝑥﷯ Step 2: Putting P’﷐𝑥﷯=0 ﷐ 𝑥 ﷐35−𝑥﷯﷮4﷯﷐70−7𝑥﷯=0 Hence 𝑥 = 0 , 10 , 35 are Critical Points But, If We Take 𝑥 = 0 Product will be 0 So, x = 0 is not possible If x = 35 𝑦 = 35 – 35 = 35 – 35 = 0 So, product will be 0 So, x = 35 is not possible Hence only critical point is 𝑥=10 Step 3: Finding P’’﷐𝑥﷯ P’﷐𝑥﷯=𝑥﷐﷐35−𝑥﷯﷮4﷯﷐70−7𝑥﷯ P’﷐𝑥﷯=﷐﷐35−𝑥﷯﷮4﷯﷐70𝑥−7﷐𝑥﷮2﷯﷯ P’’﷐𝑥﷯=﷐𝑑﷐﷐35 − 𝑥﷯﷮4﷯﷮𝑑𝑥﷯. ﷐70𝑥−7﷐𝑥﷮2﷯﷯+﷐𝑑﷐70𝑥 − 7﷐𝑥﷮2﷯﷯﷮𝑑𝑥﷯﷐﷐35−𝑥﷯﷮4﷯ =4﷐﷐35−𝑥﷯﷮3﷯.﷐𝑑﷐35 − 𝑥﷯﷮𝑑𝑥﷯. ﷐70𝑥−7﷐𝑥﷮2﷯﷯+﷐70−14𝑥﷯﷐﷐35−𝑥﷯﷮4﷯ =4﷐﷐35−𝑥﷯﷮3﷯﷐0−1﷯﷐70𝑥−7﷐𝑥﷮2﷯﷯+﷐70−14𝑥﷯﷐﷐35−𝑥﷯﷮4﷯ =−4﷐﷐35−𝑥﷯﷮3﷯﷐70𝑥−7﷐𝑥﷮2﷯﷯+﷐70−14𝑥﷯﷐﷐35−𝑥﷯﷮4﷯ Putting 𝑥 = 10 P’’﷐𝑥﷯ at 𝑥 = 10 =−4﷐﷐35−10﷯﷮3﷯﷐70﷐10﷯−7﷐﷐10﷯﷮2﷯﷯+﷐70−14﷐10﷯﷯﷐﷐35−10﷯﷮4﷯ =−4﷐﷐25﷯﷮3﷯﷐700−700﷯+﷐70−140﷯﷐﷐25﷯﷮4﷯ =−4﷐﷐25﷯﷮3﷯﷐0﷯+﷐−70﷯﷐﷐25﷯﷮4﷯ =0−70﷐﷐25﷯﷮4﷯ =−70﷐﷐25﷯﷮4﷯ < 0 Thus P’’﷐𝑥﷯<0 when 𝑥 = 10 ⇒ P is maximum when 𝑥 = 10 Thus, when 𝑥 = 10 𝑦 = 35 – 𝑥= 35 −10=25 Hence 𝒙 = 10 & 𝒚 = 25 Ex 6.5,15 (Method 2) Find two positive numbers 𝑥 and 𝑦 such that their sum is 35 and the product 𝑥2 𝑦5 is a maximum. Given two number are 𝑥 & 𝑦 Such that 𝑥 + 𝑦 = 35 𝑦 = 35 – 𝑥 Let P = 𝑥2 𝑦5 We need to maximise P Step 1: Finding P’﷐𝑥﷯ P﷐𝑥﷯=﷐𝑥﷮2﷯﷐𝑦﷮5﷯ P﷐𝑥﷯=﷐𝑥﷮2﷯﷐﷐35−𝑥﷯﷮5﷯ P’﷐𝑥﷯=﷐𝑑﷐﷐𝑥﷮2﷯﷐﷐35 − 𝑥﷯﷮5﷯﷯﷮𝑑𝑥﷯ P’﷐𝑥﷯=﷐𝑑﷐﷐𝑥﷮2﷯﷯﷮𝑑𝑥﷯ . ﷐﷐35−𝑥﷯﷮5﷯+﷐𝑑﷐﷐35 − 𝑥﷯﷮5﷯﷮𝑑𝑥﷯ . ﷐𝑥﷮2﷯ =2𝑥 .﷐﷐35−𝑥﷯﷮5﷯+﷐5﷐35−𝑥﷯﷮4﷯ .﷐𝑑﷐35 − 𝑥﷯﷮𝑑𝑥﷯ . ﷐𝑥﷮2﷯ =2𝑥 .﷐﷐35−𝑥﷯﷮5﷯+﷐5﷐35−𝑥﷯﷮4﷯ . ﷐0−1﷯﷐﷐𝑥﷮2﷯﷯ =2𝑥 .﷐﷐35−𝑥﷯﷮5﷯+﷐5﷐35−𝑥﷯﷮4﷯ ﷐−﷐𝑥﷮2﷯﷯ =2𝑥 ﷐﷐35−𝑥﷯﷮5﷯−﷐5﷐𝑥﷮2﷯ ﷐35−𝑥﷯﷮4﷯ = ﷐ 𝑥 ﷐35−𝑥﷯﷮4﷯﷐2﷐35−𝑥﷯−5𝑥﷯ = ﷐ 𝑥 ﷐35−𝑥﷯﷮4﷯﷐70−7𝑥﷯ Step 2: Putting P’﷐𝑥﷯=0 ﷐ 𝑥 ﷐35−𝑥﷯﷮4﷯﷐70−7𝑥﷯=0 ﷐ 𝑥 ﷐35−𝑥﷯﷮4﷯﷐70−7𝑥﷯=0 Hence 𝑥 = 0 , 10 , 35 are Critical Points But, If We Take 𝑥 = 0 Product will be 0 So, x = 0 is not possible If x = 35 𝑦 = 35 – 35 = 35 – 35 = 0 So, product will be 0 So, x = 35 is not possible Hence only critical point is 𝑥=10 Step 3: ⇒ 𝑥 = 10 is point of maxima P﷐𝑥﷯ is maximum at 𝑥 = 10 Thus, when 𝑥 = 10 𝑦 = 35 – 𝑥= 35 −10=25 Hence 𝒙 = 10 & 𝒚 = 25

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