Question 954660
Let the two numbers be X and Y.
You want to maximize {{{Z=X*Y}}} with {{{2X+3Y=60}}}.
From the second equation,
{{{2X=60-3Y}}}
{{{X=30-(3/2)Y}}}
Substitute into the product,
{{{Z=(30-(3/2)Y)Y}}}
Now the product is the function of only one variable.
You can either take the derivative and find the extrema or since it's a quadratic, you can convert it to vertex form to find the maximum.
{{{Z=-(3/2)Y^2+30Y}}}
{{{Z=-(3/2)(Y^2-20Y)}}}
{{{Z=-(3/2)(Y^2-20Y+100)+(3/2)(100)}}}
{{{Z=-(3/2)(Y-10)^2+150}}}
In vertex form, the parabola opens downwards and the vertex value is the maximum.
The maximum {{{Z=150}}} occurs when {{{Y=10}}}
So then,
{{{X=30-(3/2)(10)}}}
{{{X=30-15}}}
{{{X=15}}}