# Ex 6.3,15 - Chapter 6 Class 12 Application of Derivatives

Last updated at April 8, 2024 by Teachoo

Ex 6.3

Ex 6.3, 1 (i)
Important

Ex 6.3, 1 (ii)

Ex 6.3, 1 (iii) Important

Ex 6.3, 1 (iv)

Ex 6.3, 2 (i)

Ex 6.3, 2 (ii) Important

Ex 6.3, 2 (iii)

Ex 6.3, 2 (iv) Important

Ex 6.3, 2 (v) Important

Ex 6.3, 3 (i)

Ex 6.3, 3 (ii)

Ex 6.3, 3 (iii)

Ex 6.3, 3 (iv) Important

Ex 6.3, 3 (v)

Ex 6.3, 3 (vi)

Ex 6.3, 3 (vii) Important

Ex 6.3, 3 (viii)

Ex 6.3, 4 (i)

Ex 6.3, 4 (ii) Important

Ex 6.3, 4 (iii)

Ex 6.3, 5 (i)

Ex 6.3, 5 (ii)

Ex 6.3, 5 (iii) Important

Ex 6.3, 5 (iv)

Ex 6.3,6

Ex 6.3,7 Important

Ex 6.3,8

Ex 6.3,9 Important

Ex 6.3,10

Ex 6.3,11 Important

Ex 6.3,12 Important

Ex 6.3,13

Ex 6.3,14 Important

Ex 6.3,15 Important You are here

Ex 6.3,16

Ex 6.3,17

Ex 6.3,18 Important

Ex 6.3,19 Important

Ex 6.3, 20 Important

Ex 6.3,21

Ex 6.3,22 Important

Ex 6.3,23 Important

Ex 6.3,24 Important

Ex 6.3,25 Important

Ex 6.3, 26 Important

Ex 6.3, 27 (MCQ)

Ex 6.3,28 (MCQ) Important

Ex 6.3,29 (MCQ)

Last updated at April 8, 2024 by Teachoo

Ex 6.3, 15 (Method 1) Find two positive numbers š„ and š¦ such that their sum is 35 and the product š„2 š¦5 is a maximum. Given two number are š„ & š¦ Such that š„ + š¦ = 35 š¦ = 35 ā š„ Let P = š„2 š¦5 We need to maximize P Finding Pā(š) P(š„)=š„^2 š¦^5 P(š„)=š„^2 (35āš„)^5 Pā(š„)=š(š„^2 (35 ā š„)^5 )/šš„ Pā(š„)=š(š„^2 )/šš„ . (35āš„)^5+(š(35 ā š„)^5)/šš„ . š„^2 =2š„ .(35āš„)^5+ć5(35āš„)ć^4 .š(35 ā š„)/šš„ . š„^2 =2š„ .(35āš„)^5+ć5(35āš„)ć^4 . (0ā1)(š„^2 ) =2š„ .(35āš„)^5+ć5(35āš„)ć^4 (āš„^2 ) =2š„ (35āš„)^5āć5š„^2 (35āš„)ć^4 = ć š„ (35āš„)ć^4 [2(35āš„)ā5š„] = ć š„ (35āš„)ć^4 (70ā7š„) Putting Pā(š)=š ć š„ (35āš„)ć^4 (70ā7š„)=0 Hence š„ = 0 , 10 , 35 are Critical Points But, If we Take š„ = 0 Product will be 0 So, x = 0 is not possible If x = 35 š¦ = 35 ā 35 = 35 ā 35 = 0 So, product will be 0 So, x = 35 is not possible Hence only critical point is š„=10 Finding Pāā(š) Pā(š„)=š„(35āš„)^4 (70ā7š„) Pā(š„)=(35āš„)^4 (70š„ā7š„^2 ) Pāā(š„)=(š(35 ā š„)^4)/šš„. (70š„ā7š„^2 )+š(70š„ ā 7š„^2 )/šš„ (35āš„)^4 =4(35āš„)^3.š(35 ā š„)/šš„. (70š„ā7š„^2 )+(70ā14š„) (35āš„)^4 =4(35āš„)^3 (0ā1)(70š„ā7š„^2 )+(70ā14š„) (35āš„)^4 =ā4(35āš„)^3 (70š„ā7š„^2 )+(70ā14š„) (35āš„)^4 Putting š„ = 10 in Pāā(x) Pāā(š„) = ā4(35āš„)^3 (70š„ā7š„^2 )+(70ā14š„) (35āš„)^4 =ā4(35ā10)^3 (70(10)ā7(10)^2 )+(70ā14(10)) (35ā10)^4 =ā4(25)^3 (700ā700)+(70ā140) (25)^4 =ā4(25)^3 (0)+(ā70) (25)^4 =0ā70(25)^4 =ā70(25)^4 < 0 Thus, Pāā(š„)<0 when š„ = 10 ā“ P is maximum when š„ = 10 Thus, when š„ = 10 š¦ = 35 ā š„= 35 ā10=25 Hence š = 10 & š = 25 Ex 6.3, 15 (Method 2) Find two positive numbers š„ and š¦ such that their sum is 35 and the product š„2 š¦5 is a maximum. Given two number are š„ & š¦ Such that š„ + š¦ = 35 š¦ = 35 ā š„ Let P = š„2 š¦5 We need to maximise P Finding Pā(š) P(š„)=š„^2 š¦^5 P(š„)=š„^2 (35āš„)^5 Pā(š„)=š(š„^2 (35 ā š„)^5 )/šš„ Pā(š„)=š(š„^2 )/šš„ . (35āš„)^5+(š(35 ā š„)^5)/šš„ . š„^2 =2š„ .(35āš„)^5+ć5(35āš„)ć^4 .š(35 ā š„)/šš„ . š„^2 =2š„ .(35āš„)^5+ć5(35āš„)ć^4 . (0ā1)(š„^2 ) =2š„ .(35āš„)^5+ć5(35āš„)ć^4 (āš„^2 ) =2š„ (35āš„)^5āć5š„^2 (35āš„)ć^4 = ć š„ (35āš„)ć^4 [2(35āš„)ā5š„] = ć š„ (35āš„)ć^4 (70ā7š„) Putting Pā(š)=š ćš„ (35āš„)ć^4 (70ā7š„)=0 ćš„ (35āš„)ć^4 (70ā7š„)=0 Hence š„ = 0 , 10 , 35 are Critical Points But, If We Take š„ = 0 Product will be 0 So, x = 0 is not possible If x = 35 š¦ = 35 ā 35 = 35 ā 35 = 0 So, product will be 0 So, x = 35 is not possible Hence only critical point is š„=10 ā“ š„ = 10 is point of maxima P(š„) is maximum at š„ = 10 Thus, when š„ = 10 š¦ = 35 ā š„= 35 ā10=25 Hence š = 10 & š = 25