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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  • $\begingroup$ 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. $\endgroup$ – mfl Jan 28 '16 at 18:53
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    $\begingroup$ I think you confused bottom and the top here, because $y = -x^2$ never goes beyond (higher?) $x$ axis. $\endgroup$ – Kaster Jan 28 '16 at 18:53
  • $\begingroup$ I don't understand. I'm having trouble thinking about where the rectangle could possibly be. $\endgroup$ – Karl Jan 28 '16 at 18:54
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    $\begingroup$ Draw a picture. There is probably a typo, I would guess the parabola is $A-x^2$ for some positive constant $A$. $\endgroup$ – André Nicolas Jan 28 '16 at 18:56
  • $\begingroup$ @AndréNicolas that would make a lot of sense. $\endgroup$ – 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.

This is an optimization problem, because it has the word "maximum" in it. But first, you have to setup an equation to generate the area before you find the maximum of the area.

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 $a = 2Cx - 2x^3$.

Now, if you take the derivative, you can use the derivative to find the maximum value of this function.


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