Question 310349
<pre><font size = 4 color = "indigo"><b>

I think you need to draw a graph to understand what is going on:

{{{drawing(400*4/7,400,-6,6,-3,18,

graph(400*4/7,400,-6,6,-3,18,16-x^2),

green(rectangle(-4/sqrt(3),0,4/sqrt(3),16-16/3)),

green(line(  -4/sqrt(3),16-16/3, 4/sqrt(3),16-16/3)),

green(line(  -4/sqrt(3),0, 4/sqrt(3),0)),
locate(2.5,11,"(x,y)"), locate(1.6,5.8,y),
locate(1.1,1,x), locate(-1.7,1,x)
 )}}}

The base of the green rectangle is 2x, because
the base extends from -x to +x on the x axis,
and that is a distance of 2x.

The height of the green rectangle is y, so

The area of the green rectangle is

Area = base * height or

{{{A = (2x)*y}}}

Since y and x are related by the equation of the parabola,

we can substitute {{{(16-x^2)}}} for {{{y}}}

{{{A = (2x)*(16-x^2)}}}
{{{A = 32x-2x^3}}}
{{{(dA)/(dx)=32-6x^2}}}

Set {{{(dA)/(dx)=32-6x^2=0}}}

{{{32-6x^2=0}}}
{{{16-3x^2=0}}}
{{{16=3x^2}}}
{{{16/3=x^2}}}
{{{""+-sqrt(16/3)=x}}}

Use the positive value

{{{x=sqrt(16)/sqrt(3)=4/sqrt(3)=(4*sqrt(3))/(sqrt(3)*sqrt(3))=(4sqrt(3))/3}}}

Using the second derivative test,

{{{(dA)/(dx)=32-6x^2}}}
{{{(d^2A)/(dx^2)=-12x}}}, which is negative when {{{x=(4sqrt(3))/3}}},
which proves that A is a maximum when {{{x=(4sqrt(3))/3}}}.

Therefore, the base of the largest rectangle{{{""=2x=2*((4sqrt(3))/3)=(8sqrt(3))/3}}}, 
and its height is{{{""=y = 16-x^2=16-16/3=48/3-16/3=32/3}}}.

Edwin</pre>