Question 970464
{{{perimeter=2(length+width)}}}
Let us define some variables
{{{L}}}= length of the rectangle, in cm.
{{{W}}}= width of the rectangle, in cm.
{{{A}}}=area of the rectangle in square centimeters.
The way {{{l}}} and {{{W}}} were defined,
{{{L>=W}}} , and when {{{L=W}}} we have that special type of rectangle that we call a square.
From geometry (and real life common sense) we know that
{{{perimeter}}}{{{"(in"}}}{{{"cm)" =2(L+W)}}} and {{{A=L*W}}} .
From the problem, we know that
{{{perimeter=100}}}{{{cm}}} .
From all that we know,
{{{2(L+W)=100}}} ---> {{{L+W=100/2}}} ---> {{{L+W=50}}} ---> {{{L=50-W}}} ,
and plugging the expression {{{50-W}}} for {{{L}}} in {{{A=L*W}}} ,
we get {{{A}}} as a function of {{{W}}} :
{{{A=(50-W)W}}} <---> {{{A=50W-W^2}}} .
That tells us that the area of the rectangle is a quadratic function of {{{W}}} .
From a memorized formula,
or from looking a the function transformed using algebra as shown below,
we realize that the function  has a maximum for {{{W=25}}} ,
when {{{A=25*25=625}}} .
To both sides of {{{W=25}}} ,, the farther we go from {{{W=25}}} ,
the smaller the value for {{{A}}} .
Of course, we defined {{{W}}} , {{{L}}} , and {{{A}}} to make sense with the "real life" situation of the problem, so they are all positive, for starters.
Also since {{{L=50-W}}} and {{{L>=W}}} ,
{{{50-W>=W}}} ---> {{{50>=2W}}} ---> {{{50/2>=W}}} ---> {{{25>=W}}} , so the function is defined for {{{0<W<=25}}} ,
and we get only a piece of the quadratic function,
increasing from {{{W=almost}}}{{{0}}} to a maximum for {{{W=25}}} ,
when {{{A=25*25=625}}} .
{{{A=50W-W^2}}}<--->{{{A=-W^2+50-625+625}}}<--->{{{A=-(W^2-50W+625)+625}}}<---> {{{A=-(W-25)^2+625}}}
If you need an area of 500 square centimeters, you would get it when
{{{500=50W-W^2}}} , so you solve that to find the minimum width.
Or, you say that you want {{{A>=500}}} and use the expression
{{{A=-(W-25)^2+625}}}<--->{{{A=-(25-W)^2+625}}} to get
{{{-(25-W)^2+625>=500}}} ---> {{{625-500>=(25-W)^2}}} ---> {{{125>=(W-25)^2}}} ---> {{{sqrt(125)>=25-W}}} (because we know that {{{W<=25}}}<--->{{{25-W>=0}}} ) .
Then ,
{{{sqrt(125)>=25-W}} ---> {{{5sqrt(5)>=25-W}} ---> {{{highlight(W>=25-5sqrt(5))}}}
and {{{25-5sqrt(5)=about22.76}}} .