Ex 6.3

Chapter 6 Class 12 Application of Derivatives
Serial order wise

### Transcript

Ex 6.3, 14 (Method 1) Find two positive numbers š„ and y such that š„ + š¦ = 60 and š„š¦3 is maximum. Given two number š„ and y, such that š„ + š¦ = 60 š¦=60āš„ Let P = š„š¦3 We need to maximize P Now, P = š„š¦3 Putting value of y from (1) P = š„(60āš„)3 Finding Pā(x) P = š„(60āš„)^3 Diff w.r.t š„ šš/šš„=š(š„(60 ā š„)^3 )/šš„ šš/šš„=š(š„)/šš„ (60āš„)^3+(š(60 ā š„)^3)/šš„ . š„ =(60āš„)^3+ć3(60āš„)ć^2 . (0ā1) . š„ =(60āš„)^3ā3š„(60āš„)^2 =(60āš„)^2 (60āš„)ā3š„(60āš„)^2 =(60āš„)^2 [(60āš„)ā3š„] =(60āš„)^2 [60ā4š„] Putting šš·/šš=š (60āš„)^2 (60ā4š„)=0 So, x = 60 & x = 60/4 = 15 But, If š„=60, š¦= 60 ā š„ = 60 ā 60 = 0 Which is not possible Hence, š„= 15 is only critical point. Finding Pāā (š) Pāā (š„)=š((60 ā š„)^2 (60 ā 4š„))/šš„ Pāā (š„)=(š(60 ā š„)^2)/šš„ . (60ā4š„)+š(60 ā 4š„)/šš„ (60āš„)^2 = 2(60āš„) .(0ā1)(60ā4š„)ā4(60āš„)^2 = ā2(60āš„) . (60ā4š„)ā4(60āš„)^2 = ā2(60āš„)[(60ā4š„)+2(60āš„)] = ā2(60āš„)[(60ā4š„)+120ā2š„] = ā2(60āš„)(180ā6š„) At š = 15 Pāā(15)=ā2(60ā15)(180ā6(15)) =ā90 Ć90 =ā8100 < 0 ā“ Pāā(š„)<0 at š„ = 15 Hence š„š¦3 is Maximum when š„ = 15 Thus, when š„ = 15 š¦ =60 ā š„=60 ā15=45 Hence, numbers are 15 & 45