Question 854501
It's a good 1st step to draw the parabola
and also a horizontal line which represents
both the point on the y-axis ( 0, y ) and
also the intersection of this line with the 
parabola ( x, y ).
Remember that the horizontal line can be
ANY horizontal line
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{{{ graph( 400, 400, -10, 10, -10, 40, 24, -x^2 + 36 ) }}}
So far,  I have the points:
( 0, 0 )
( 0, y )
( x, y )
Now I can add the final point ( x1, 0 )
Both x and y are positive
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{{{ A = x*y }}}
{{{ A = x*( -x^2 + 36 ) }}}
{{{ A = -x^3 + 36x }}}
I can plot this on top of the previous plot
{{{ graph( 400, 400, -10, 10, -10, 100, 18, -x^2 + 36, -x^3 + 36x ) }}}
The domain is {{{ 0 }}} to {{{ 6 }}}
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Are you in a calculus class? That's the only way I know to 
find the peak of 3rd degree equation
{{{ A = -x^3 + 36x }}}
{{{ A' = -3x^2 + 36 }}}
{{{ A' = 0 }}} ( slope = 0 )
{{{ -3x^2 + 36 = 0 }}}
{{{ 3x^2 = 36 }}}
{{{ x^2 = 12 }}}
{{{ x[max] = 2*sqrt(3) }}}
This is the value of {{{x}}} for which {{{ A }}} is largest
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and now find {{{ y[max] }}}
{{{ y[max] = -x^2 + 36 }}}
{{{ y[max] = -( 2*sqrt(3) )^2 + 36 }}}
{{{ y[max] = -4*3 + 36 }}}
{{{ y[max] = 36 - 12 }}}
{{{ y[max] = 24 }}}
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Hope this all makes sense- if you do it with a
non-calculus method, you should get the
same answer