I have a questions asking for the dimensions of the rectangle with the largest area that has two bottom corners on the x axis and two top corners on the curve $y=12-x^2$.

I have plotted the curve and found it is a symmetrical parabola with a vertex of $x=0, y=12$.

It intersects the $x$ axis at $-2\sqrt3$ and $2\sqrt3$.

My thinking is that if I find when the derivative of the (area under the curve, minus the area inside the square) = 0, then I can determine what values make it a minimum.

I also thought that I could half the parabola and work with one side since it is symmetrical, then double those values at the end.

So the area under the curve in the positive x axis = $∫_0^{2\sqrt3}12-x^2 dx$

My problem is that I can't define area of the rectangle, or the sides. Can anyone give me any pointers?

  • 1
    $\begingroup$ You don't need to find the area under the curve. Just let the top right corner of your rectangle be the coordinates $(x,y) = (x, 12-x^2)$. Hopefully you can go from there. $\endgroup$ – matthras Oct 29 '13 at 10:49
  • $\begingroup$ Yeah I'm not seeing where that came from that's all $\endgroup$ – user88720 Oct 29 '13 at 11:01

As you say, let us take into account the symetry. Let us put the two bottom corner at (a,0) and (-a,0). The top corners will be at (-a,b) and (a,b). But the top corners are also on the parabola; this means that b=12-a^2. Then, the area is 2ab = 2 a (12 - a^2) and you want this area be maximum. Are you able to continue with this ? If not, just post a message to me.

  • $\begingroup$ I follow apart from the bit about $b=12-a^2$ I'm not sure where that came from $\endgroup$ – user88720 Oct 29 '13 at 10:52
  • $\begingroup$ Since the top corner is on the parabola given by y = 12 - x^2 and the bottom corner is at (a,0) then b = 12 - a^2. Is this clearer ? Plot your function, add two points (-a,0) and (a,0) (as you noticed, a < 2 Sqrt[3]); draw two vertical lines until they intersect the parabola. The y intersect is b. $\endgroup$ – Claude Leibovici Oct 29 '13 at 11:03
  • $\begingroup$ Ah yep I get it now. I wasn't making that relation between the box height and width to the curve. All good. Dimensions should be 4 along the x axis and 8 along the y. Cheers $\endgroup$ – user88720 Oct 29 '13 at 11:17
  • $\begingroup$ You got it ! Simple, isn't. I am happy to have been able to help you $\endgroup$ – Claude Leibovici Oct 29 '13 at 17:04

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