Question 1098923
*[illustration trz2.png].
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{{{A=XY+(1/2)(XY)=(3/2)XY}}}
and
{{{2X+Y+X=1200}}}
{{{3X+Y=1200}}}
{{{Y=1200-3X}}}
Substituting to get the area as a function of only one variable,
{{{A=(3/2)X(1200-3X)}}}
{{{A=(3/2)(-3X^2+1200X)}}}
Complete the square to get the area to vertex form,
{{{A=-(9/2)(X^2-400X)}}}
{{{A=-(9/2)(X^2-400X+200^2)+(9/2)(200^2)}}}
{{{A=-(9/2)(X-200)^2+(9)(200*100)}}}
{{{A=-(9/2)(X-200)^2+180000}}}
Vertex solution is the maximum.
{{{A[max]=180000}}}{{{m^2}}} when {{{X=200}}}{{{m}}}