Question 823745
The parabola looks like this
{{{graph(300,450,-5,5,-2,13,10-x^2)}}} , with x-intercepts at {{{sqrt(10)}}} and {{{-sqrt(10)}}} .
( {{{sqrt(10)}}} = approx. {{{3.16}}} ).
The rectangle in the first quadrant looks like this:
{{{drawing(300,300,-5,5,-2,8,
grid(0),
red(circle(1.83,6.67,0.1)),
locate(1.9,7,P(x,y)),
rectangle(0,0,1.83,20/3),
locate(0.85,6.7,x),
locate(1.9,3.5,y=10-x^2)
)}}} We want a formula to calculate the area of that rectangle, and we want to find the maximum area.
 
a.) {{{A(x)=x*y=x(10-x^2)=10x-x^3}}} for {{{0<x<sqrt(10)}}} .
We have to put that restriction on the domain, because {{{P(x,y)}}} must be in the first quadrant.
So we can write the function as
{{{highlight(A(x)=-x^3+10x)}}}  for {{{highlight(0<x<sqrt(10))}}} .
It is a polynomial function.
Fully factored, it can be written as
{{{A(x)=-x(x+sqrt(10))(x-sqrt(10))}}}
If the domain were not restricted, we would say that it has zeros at {{{x=-sqrt(10)}}, {{{x=0}}} and {{{x=sqrt(10)}}} .
We would figure out that it is positive and decreasing for {{{x<-sqrt(10)}},
where {{{x<0}}} , {{{x+sqrt(10)<0}}} and {{{x+sqrt(10)<0}}} .
It is negative for {{{-sqrt(10)<x<0}}} and going through a minimum in that interval.
For {{{0<x<sqrt(10)}}} , {{{A(x) is positive and goes through a maximum.
Finally, for {{{x>sqrt(10)}}} , the function is negative and decreasing.
Here's the graph {{{graph(200,300,-5,5,-15,15,-x^3+10x)}}}
andpositive for {{{x=0}}} and {{{x>sqrt(10)}}} .
 
b.) The only way that I know to find the maximum area of such a rectangle requires using calculus and finding the derivative of {{{A(x)}}}
{{{dA/dt=-3x^2+10}}}
The zeros of that derivative show us the location of the minimum and maximum.
{{{-3x^2+10=0}}}
{{{10=3x^2}}}
{{{10/3=x^2}}}
The solutions are:
{{{x=-sqrt(10/3)=-sqrt(30)/3}}} or {{{x=sqrt(10/3)=sqrt(30)/3}}}
The maximum occurs at {{{x=sqrt(30)/3}}} , where {{{x^2=10/3}}}
Substituting those values into 
{{{A(x)=x(10-x^2)}}} we can easily calculate
{{{A[max]=(sqrt(30)/3)(10-10/3)=(sqrt(30)/3)(20/3)=highlight(20sqrt(30)/9)}}}
The approximate value is {{{highlight(12.17)}}} .