Math
posted by Mary on .
A rectangle is inscribed with its base on the xaxis and its upper corners on the parabola y=9x^2. What are the dimensions of such a rectangle with the greatest possible area?

let the point of contact in the first quadrant be (x,y)
then the base of the rectangle is 2x and its height is y
Area = 2xy
= 2x(9x^2)
= 18x  2x^2
d(area)/dx = 18  4x
so 184x=0
x = ....
take it from here, let me know what you got 
change
<< = 18x  2x^2 >> to
= 18x  2x^3
then d(area)/dx = 18  6x^2
= 0 for a max/min of area
6x^2 = 18
x^2 = 3
x = ±√3
and y = 6
same result as Damon 
Well, this is an upside down parabola (sheds water) with x intercepts at x = 3 and x = +3 and vertex at (0,9)
so lets do it just for positive x and double the base at the end.
Area = x y
where y = 9x^2
A = 9x  x^3
dA/dx = 9  3 x^2
that is zero when x^2 = 3
or x = +/ sqrt 3 use + sqrt 3
so y = 9  3 = 6
Now we double the base because we only did the right half
base = 2 sqrt 3
height = 6 
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