Question 898964
Try solving this in general.
A for area, p for perimeter.
Dimensions x and y.


Rectangle.
{{{xy=A}}} and {{{2x+2y=p}}}.
Assume A and p are known.
{{{2y=p-2x}}}
{{{y=(p-2x)/2}}}
{{{y=p/2-x}}}------substitute this into the area equation.
-
{{{x(p/2-x)=A}}}
{{{(P/2)x-x^2-A=0}}}
{{{-x^2+(p/2)x-A=0}}} and multiply both members by -1.
{{{highlight_green(x^2-(p/2)x+A=0)}}}
which, if you substitute for the values of p and A now, you might find that the
quadratic expression is factorable; but you do not always have this condition.


{{{x=(p/2+- sqrt((p/2)^2-4*A))/2}}}
You most likely want the PLUS square root form;
{{{highlight(x=(p/2+sqrt((p/2)^2-4A))/2)}}}


Use the area equation to solve for y.
{{{y=A/x}}}
{{{A/((p/2+sqrt((p/2)^2-4A))/2)}}}
{{{highlight(y=(2A)/(p/2+sqrt((p/2)^2-4A)))}}}


You can substitute for A and p early and avoid pure symbolic form, but if the quadratic expression is
not factorable, the solution in pure symbolic form will work for all problems of this general kind.