a+b=15
a^2b^3=m
Find the number whose sum is 15. If the product of the square of one by the cube of the other is to be maximum
3 answers
a+b=15
a^2b^3=m
a=15-b
(15-b)^2b^3=m
(225-30b+b^2)b^3=m
225b^3-30b^4+b^5=m
dm/db=675b^2-120b^3+5b^4
at dm/db=0
675b^2-120b^3+5b^4=0
b^2(675-120b+5b^2)=0
divide through by b^2
5b^2-120b+675=0
divide through by 5
b^2-24b+135=0
plz solve quadracticly and sub into the first equation to find a
a^2b^3=m
a=15-b
(15-b)^2b^3=m
(225-30b+b^2)b^3=m
225b^3-30b^4+b^5=m
dm/db=675b^2-120b^3+5b^4
at dm/db=0
675b^2-120b^3+5b^4=0
b^2(675-120b+5b^2)=0
divide through by b^2
5b^2-120b+675=0
divide through by 5
b^2-24b+135=0
plz solve quadracticly and sub into the first equation to find a
b^2-24b+135 = (b-9)(b-15)
So, the extrema are at b=9 or b=15.
Naturally, at b=15, a=0, so the product is a minimum
At b=9, a=6, and the product is
6^2*9^3 = 26244
So, the extrema are at b=9 or b=15.
Naturally, at b=15, a=0, so the product is a minimum
At b=9, a=6, and the product is
6^2*9^3 = 26244