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Here is the problem:

A rectangle has its base on the x-axis and its upper two vertices on the parabola y=12-x^2 What is the largest area the rectangle can have, and what are its dimensions?

Well, I don't really know where to start. My initial idea was to find inflection points because I figured that is where the vertices would be, but there are no inflection points because it is a parabola...

Then I though about finding where the derivative and the parabola cross, found it, but I don't know how that will help me.

I really don't know where to start.

Any help is appreciated, thanks.

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    You should be able to express the area of the inscribed rectangle in terms of $x$. Differentiate that, and find the values of $x$ that make it zero, and check that your results do correspond to a maximum.2011-12-21
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    Okay, so: xy=A, X(12-x^2)=A, -3x^2 + 12 = 0, x = 2, y = 8, Area = 16.2011-12-21
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    @user21589: Almost there! Area is $2x(12-x^2)$, base goes from $-x$ to $x$. If you have doubts, picture will take care of them. You found the area of the first quadrant part of the best rectangle.2011-12-21

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