Question 1125335

let first number be {{{x}}}, and second number {{{y}}}

given:

{{{x+2y=120}}}...solve for {{{y}}}
{{{2y=120-x}}}
{{{y=60-x/2}}}...eq.1


the product is a maximum:

{{{P=xy}}}...substitute {{{y}}} from eq.1

{{{P=x(60-x/2)}}}

{{{P= 60x-x^2/2}}}

{{{P= -(1/2)x^2+60x}}}...............find first derivative (if you are familiar with it)

{{{P}}}'= {{{-2(1/2)x+60}}}

{{{P}}}'= {{{-x+60}}}

max is at {{{P}}}'={{{0}}}

{{{0= -x+60}}}

{{{x=60}}}

go to {{{x+2y=120}}}, find {{{y}}}

{{{60+2y=120}}}
{{{2y=120-60}}}
{{{2y=60}}}
{{{y=30}}}


your numbers are: {{{60}}} and {{{30}}}