Question 1146559
<pre>
Take your value for part (a) which is

{{{matrix(1,3,A(x),""="",4x*sqrt(36-x^2))}}}

We could take the derivative of that and set that derivative equal
to 0.  However that would be very time-consuming.  It would be much
easier to square both sides and find the value of x that maximizes
the value of the SQUARE of the area A, because if A² is the largest 
it can be, then A is also the largest it can b.

{{{matrix(1,4,Let,S(x),""="",(A(x)^"")^2)}}}

{{{matrix(11,3,
S(x),""="",(4x^""*sqrt(36-x^2))^2,
S(x),""="",16x^2(36-x^2),
S(x),""="",576x^2-16x^4,
"S'(x)",""="",1152x-64x^3,
1152x-64x^3,""="",0,
64x(18-x^2),""="",0,
64x=0,and,18-x^2=0,
x=0,and,18=x^2,
"","",sqrt(18)=x,
"","",sqrt(9*2)=x,
"","",3sqrt(2)=x  )}}}

We find the y value when x=3√3:

{{{matrix(9,2,
y,""="",
sqrt(36-x^2),""="",
sqrt(36-(3sqrt(2))^2),""="",
sqrt(36-(3^2*2)),""="",
sqrt(36-9*2),""="",
sqrt(36-18),""="",
sqrt(18),""="",
sqrt(9*2),""="",
3sqrt(2),"" )}}}


So the area is greatest when the inscribed rectangle is a square, when
the upper right hand corner is the point P(3√2, 3√2).
(The 0 value is when the area is least, when the rectangle degenerates
into just a line segment, whose area is 0 because its width is 0)

{{{drawing(400,400,-6.9,6.9,-6.9,6.9,circle(0,0,6),
graph(400,400,-6.9,6.9,-6.9,6.9),
locate(4.5,4.5,P(3sqrt(2),3sqrt(2)) ),
line(-sqrt(18),-sqrt(18),sqrt(18),-sqrt(18)),
line(-sqrt(18),sqrt(18),sqrt(18),sqrt(18)),
rectangle(-sqrt(18),-sqrt(18),sqrt(18),sqrt(18)))}}}

I think this is the graph your teacher wants.  But if he/she wants
the graph of 
{{{matrix(1,3,A(x),""="",4x*sqrt(36-x^2))}}}, 
then it is the graph below whose maximum point is (3√2, 72).  To 
plot it, you can only substitute values x=0 through 6.

{{{drawing(400,400,-.9,6.9,-2,75, 
red(line(6,0,6,10)),

graph(400,400,-.9,6.9,-2,75,4*x*sqrt(36-x^2)))}}}
 
Edwin</pre>