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?
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.