Question 329197
1.{{{X+Y=60}}}
You are trying to minimize the value of 
2.{{{Z=X^2+12Y}}}
From eq. 1,
{{{Y=60-X}}}
Substitute into eq. 2,
{{{Z=X^2+12(60-X)}}}
{{{Z=X^2-12X+720}}}
Complete the square to put the equation in vertex form,{{{y=a(x-h)^2+k}}}
Since the coefficient of {{{X^2}}} is positive, the vertex y value(k) will be the minimum.
{{{Z=X^2-12X+36+720-36}}}
{{{Z=(X-6)^2+684}}}
{{{X=6}}}
{{{Y=60-6=54}}}
.
.
.

{{{highlight(X=6)}}} and {{{highlight(Y=54)}}} would provide the minimum value of 684.