Question 1084031
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<pre>
The length of the rectangle is 2x (from -x to +x). It is "horizontal" dimension.


The height of the rectangle is f(x) = {{{8 - 2x^2}}}. It is "vertical" dimension.


The area of the rectangle is  A(x) = {{{2x*(8-2x^2)}}} = {{{16x - 4x^3}}}.


To find the maximum, take the derivative and equate it to zero:

{{{(dA(x))/(dx)}}} = 16 - 12x^2 = 0.

====>  {{{x^2}}} = {{{16/12}}}  ====>  x = {{{sqrt(16/12)}}} = {{{4/(2*sqrt(3))}}} = {{{(2*sqrt(3))/3}}}.


You calculate maximal area A(x) by substituting x = {{{(2*sqrt(3))/3}}}.
</pre>


{{{graph( 330, 330, -2.5, 5.5, -12.5, 15.5,
          16x - 4x^3
)}}}


Plot A(x) = {{{16x - 4x^3}}}