Question 785549
{{{x}}} = length of one side in meters
{{{y}}} = length of the adjacent side in meters
(one is the length  of the rectangle, and the other one is the width)
The perimeter (in meters) is
{{{2x+2y=30}}} --> {{{x+y=15}}} --> {{{y=15-x}}}
The area is
{{{area=x*y}}}
Substituting {{{15-x}}} for {{{y}}}, we get
{{{area=x(15-x)}}} <--> {{{area=15x-x^2}}} (area expressed in terms of x)
THat is a quadratic function.
It graphs as a parabola, and the vertex is a maximum.
(It is realy a portion of a parabola, because we must only define it for
{{{0<x<15}}} to have positive numbers for width odf the rectangle.
{{{area=15x-x^2}}}-->{{{area=-x^2+15x-56.25+56.25}}}-->{{{area=-(x^2-15x-56.25)+56.25}}}-->{{{area=-(x-7.5)^2+56.25}}}
The maximum area is found when {{{x=7.5}}}, and it is {{{56.25}}}.
At that point {{{y=15-7.5=7.5}}} and the rectangle is a square.
For any other value of {{{x}}}, tyhe area is less than that:
{{{(x-7.5)^2>0}}}, {{{-(x-7.5)^2<0}}}, and {{{area=-(x-7.5)^2+56.25<56.25}}}