Question 1200217
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A rectangle is inscribed with its base on the x-axis and its upper corners 
on the parabola y=9−x2. What are the dimensions of such a rectangle 
with the greatest possible area?
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<pre>
The area of any such inscribed rectangle is 

    A = A(x) = base*height = (2x)*y = 2x*(9-x^2) = 18x - 2x^3.


We want maximize the area A(x), i.e. maximize function A(x).


So, we take the derivative and equate it to zero

    A'(x) = 18 - 6x^2 = 0.


From this equation, we find

    6x^2 = 18  --->  x^2 = 18/6 = 3  --->  x = {{{sqrt(3)}}}.


The dimensions of the inscribed rectangle are  {{{2*sqrt(3)}}}  horizontally
and  {{{9 - (sqrt(3))^2}}} = 9 - 3 = 6 units vertically.


The maximum area is  xy = {{{2*6*sqrt(3)}}} = {{{12*sqrt(3)}}}  square units.
</pre>

Solved.