Asked by Abs
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
Answers
Answered by
sam
a+b=15
a^2b^3=m
a^2b^3=m
Answered by
sam
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
Answered by
Steve
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
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