3.Divide 30 into three parts such that the continued product of the fi

1.	3.Divide 30 into three parts such that the continued product of the first, thesquare of the second and the cube of the third be a maximum.Divide 120 into three parts so that the sum of the products taken two ata time shall be maximum.duet of the first square
Answer» square-1 cube-1 maximum -281+1+28=30 2)10×10^2×10^3=1000000 that's the maximum square 100 and cube 1000 40+40+40 =120 and some of the part 40 x 40 =1600 so correct answer is 40 2.)dividing 30 into three parts=10 ,10,10 101001000= 1000000 2.) division 30.into three parts= 10' 10' 10.101001000=1000000 1000000 is answer.......

3.Divide 30 into three parts such that the continued product of the first, thesquare of the second and the cube of the third be a maximum.Divide 120 into three parts so that the sum of the products taken two ata time shall be maximum.duet of the first square

Answer»

square-1 cube-1 maximum -281+1+28=30

2)10×10^2×10^3=1000000 that's the maximum

square 100 and cube 1000

40+40+40 =120 and some of the part 40 x 40 =1600 so correct answer is 40

2.)dividing 30 into three parts=10 ,10,10 10*100*1000= 1000000

2.) division 30.into three parts= 10' 10' 10.10*100*1000=1000000

1000000 is answer.......