Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives
Last updated at April 16, 2024 by Teachoo
Ex 6.3
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 You are here
Ex 6.3,15 Important
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 16, 2024 by Teachoo
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¦=60āš„ Let P = š„š¦3 We need to maximize P Now, P = š„š¦3 Putting value of y from (1) P = š„(60āš„)3 Finding Pā(x) P = š„(60āš„)^3 Diff w.r.t š„ šš/šš„=š(š„(60 ā š„)^3 )/šš„ šš/šš„=š(š„)/šš„ (60āš„)^3+(š(60 ā š„)^3)/šš„ . š„ =(60āš„)^3+ć3(60āš„)ć^2 . (0ā1) . š„ =(60āš„)^3ā3š„(60āš„)^2 =(60āš„)^2 (60āš„)ā3š„(60āš„)^2 =(60āš„)^2 [(60āš„)ā3š„] =(60āš„)^2 [60ā4š„] Putting š š·/š š=š (60āš„)^2 (60ā4š„)=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 Pāā (š) Pāā (š„)=š((60 ā š„)^2 (60 ā 4š„))/šš„ Pāā (š„)=(š(60 ā š„)^2)/šš„ . (60ā4š„)+š(60 ā 4š„)/šš„ (60āš„)^2 = 2(60āš„) .(0ā1)(60ā4š„)ā4(60āš„)^2 = ā2(60āš„) . (60ā4š„)ā4(60āš„)^2 = ā2(60āš„)[(60ā4š„)+2(60āš„)] = ā2(60āš„)[(60ā4š„)+120ā2š„] = ā2(60āš„)(180ā6š„) At š = 15 Pāā(15)=ā2(60ā15)(180ā6(15)) =ā90 Ć90 =ā8100 < 0 ā“ Pāā(š„)<0 at š„ = 15 Hence š„š¦3 is Maximum when š„ = 15 Thus, when š„ = 15 š¦ =60 ā š„=60 ā15=45 Hence, numbers are 15 & 45