Ex 6.5,14 - Chapter 6 Class 12 Application of Derivatives

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Ex 6.5,14 (Method 1) Find two positive numbers and y such that + = 60 and 3 is maximum. Given two number and y, such that + = 60 =60 Let P = 3 We need to maximize P Now, P = 3 P = 60 3 Step 1: Finding P (x) P = 60 3 Diff w.r.t = 60 3 = 60 3 + 60 3 . = 60 3 + 3 60 2 . 0 1 . = 60 3 +3 60 2 = 60 2 60 3 = 60 2 60 4 Step 2: Putting =0 60 2 60 4 =0 So, x = 60 & x = 60 4 = 15 But, If =60 = 60 60 = 0 Which is not possible Hence = 15 is only critical point. Step 3: Finding P P = 60 2 60 4 P = 60 2 . 60 4 + 60 4 60 2 = 2 60 . 0 1 60 4 4 60 2 = 2 60 . 60 4 4 60 2 = 2 60 60 4 +2 60 = 2 60 60 4 +120 2 = 2 60 180 6 At = 15 P 15 = 2 60 15 180 6 15 = 90 90= 8100 < 0 P <0 at = 15 Hence 3 is Maximum when = 15 Thus, when = 15 =60 =60 15=45 Hence, numbers are 15 & 45 Ex 6.5,14 (Method 2) Find two positive numbers and y such that + = 60 and 3 is maximum. Given two number and y, such that + = 60 =60 Let P = 3 We need to maximize P Now, P = 3 P = 60 3 Step 1: Finding P (x) P = 60 3 Diff w.r.t = 60 3 = 60 3 + 60 3 . = 60 3 + 3 60 2 . 0 1 . = 60 3 +3 60 2 = 60 2 60 3 = 60 2 60 4 Step 2: Putting =0 60 2 60 4 =0 So, x = 60 & x = 60 4 = 15 But, If =60 = 60 60 = 0 Which is not possible Hence = 15 is only critical point. Step 3: = 15 is point of local Maxima & P is Maximum at = 15 Thus, when = 15 =60 =60 15=45 Hence, numbers are 15 & 45