The sum of two natural numbers equals 120, in which the multiplication of the square of one of them by the other number is to be as maximum as possible, how do you find the two numbers?

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Answer 1 · 2016-12-30T06:49:10

a = 80, b=40

let say the two numbers are a and b.

$a+b=120$
$b = 120-a$

let say that a is a number to be squared.
$y=a^2*b$
$y=a^2*(120-a)$
$y=120a^2-a^3$

$dy/dx = 240a-3a^2$

max or min when $dy/dx=0$
$240a-3a^2=0$
$a(240-3a)=0$

$a=0 and 80$
$b=120 and 40$

$(d^2y)/(dx^2) = 240-6a$

when a =0,
$(d^2y)/(dx^2) = 240$ . minimum

when a =80,
$(d^2y)/(dx^2) = -240$ . maximum.

the answer is a = 80 and b =40.