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

Last updated at April 16, 2024 by Teachoo

Ex 6.3

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Last updated at April 16, 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