Question 1106230
{{{A+B=60}}}
So then what is the maximum of {{{Z=AB}}}?
You can substitute from above,
{{{A=60-B}}}
So,
{{{Z=(60-B)B}}}
{{{Z=60B-B^2}}}
So find the maximum of Z.
Since there is a negative coefficient in front of the quadratic term, the parabola opens downward and has a maximum.
Since it's a parabola you can convert to vertex form and find the maximum and the location of the maximum.