# 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?

Dec 30, 2016

a = 80, b=40

#### Explanation:

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.
$y = {a}^{2} \cdot b$
$y = {a}^{2} \cdot \left(120 - a\right)$
$y = 120 {a}^{2} - {a}^{3}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 240 a - 3 {a}^{2}$

max or min when $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$240 a - 3 {a}^{2} = 0$
$a \left(240 - 3 a\right) = 0$

$a = 0 \mathmr{and} 80$
$b = 120 \mathmr{and} 40$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 240 - 6 a$

when a =0,
$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 240$. minimum

when a =80,
$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - 240$. maximum.

the answer is a = 80 and b =40.