# Rectangle under a parabola. [closed]

You are given that $y=-x^2$ find a show the maxium area you can create using a rectangle with its bottom edge along the $x$ axis and its top two corners lie on the curve $y=-x^2$.

Any idea?

## closed as unclear what you're asking by Silvia Ghinassi, Kaster, Yagna Patel, colormegone, user147263 Jan 28 '16 at 22:02

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• There is no such a rectangle. You can find rectangles arbitrarily large. So, are you sure your question is correct? Also, please edit the tags. It is not related to integration nor graph-theory. – mfl Jan 28 '16 at 18:53
• I think you confused bottom and the top here, because $y = -x^2$ never goes beyond (higher?) $x$ axis. – Kaster Jan 28 '16 at 18:53
• I don't understand. I'm having trouble thinking about where the rectangle could possibly be. – Karl Jan 28 '16 at 18:54
• Draw a picture. There is probably a typo, I would guess the parabola is $A-x^2$ for some positive constant $A$. – André Nicolas Jan 28 '16 at 18:56
• @AndréNicolas that would make a lot of sense. – Kaster Jan 28 '16 at 18:56

I am going to presume that the question was $y = C - x^2$, since that would make more sense.
So, first, what is the equation of the area of a rectangle? $a = length\cdot width$.
So, on this rectangle, what will be the width? Since the parabola is centered around the $x$ axis, it will be 2 times the $x$ value, or $2x$. What will be the height? It will be the value of the $y$ for the given $x$. So $a = 2x\cdot y$.
Now, we have an equation for the $y$ value, $y = C - x^2$. Therefore we can substitute: $a = 2x\cdot (C - x^2)$. This simplifies to $a = 2Cx - 2x^3$.