Ex 6.5,15 - Chapter 6 Class 12 Application of Derivatives

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Ex 6.5,15 (Method 1) Find two positive numbers and such that their sum is 35 and the product 2 5 is a maximum. Given two number are & Such that + = 35 = 35 Let P = 2 5 We need to maximise P Step 1: Finding P P = 2 5 P = 2 35 5 P = 2 35 5 P = 2 . 35 5 + 35 5 . 2 =2 . 35 5 + 5 35 4 . 35 . 2 =2 . 35 5 + 5 35 4 . 0 1 2 =2 . 35 5 + 5 35 4 2 =2 35 5 5 2 35 4 = 35 4 2 35 5 = 35 4 70 7 Step 2: Putting P =0 35 4 70 7 =0 Hence = 0 , 10 , 35 are Critical Points But, If We Take = 0 Product will be 0 So, x = 0 is not possible If x = 35 = 35 35 = 35 35 = 0 So, product will be 0 So, x = 35 is not possible Hence only critical point is =10 Step 3: Finding P P = 35 4 70 7 P = 35 4 70 7 2 P = 35 4 . 70 7 2 + 70 7 2 35 4 =4 35 3 . 35 . 70 7 2 + 70 14 35 4 =4 35 3 0 1 70 7 2 + 70 14 35 4 = 4 35 3 70 7 2 + 70 14 35 4 Putting = 10 P at = 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 <0 when = 10 P is maximum when = 10 Thus, when = 10 = 35 = 35 10=25 Hence = 10 & = 25 Ex 6.5,15 (Method 2) Find two positive numbers and such that their sum is 35 and the product 2 5 is a maximum. Given two number are & Such that + = 35 = 35 Let P = 2 5 We need to maximise P Step 1: Finding P P = 2 5 P = 2 35 5 P = 2 35 5 P = 2 . 35 5 + 35 5 . 2 =2 . 35 5 + 5 35 4 . 35 . 2 =2 . 35 5 + 5 35 4 . 0 1 2 =2 . 35 5 + 5 35 4 2 =2 35 5 5 2 35 4 = 35 4 2 35 5 = 35 4 70 7 Step 2: Putting P =0 35 4 70 7 =0 35 4 70 7 =0 Hence = 0 , 10 , 35 are Critical Points But, If We Take = 0 Product will be 0 So, x = 0 is not possible If x = 35 = 35 35 = 35 35 = 0 So, product will be 0 So, x = 35 is not possible Hence only critical point is =10 Step 3: = 10 is point of maxima P is maximum at = 10 Thus, when = 10 = 35 = 35 10=25 Hence = 10 & = 25