         1. Chapter 6 Class 12 Application of Derivatives
2. Serial order wise
3. Ex 6.5

Transcript

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

Ex 6.5 