# Optimisation of a rectangles area under a function curve

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?

• 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. – matthras Oct 29 '13 at 10:49
• Yeah I'm not seeing where that came from that's all – user88720 Oct 29 '13 at 11:01

• I follow apart from the bit about $b=12-a^2$ I'm not sure where that came from – user88720 Oct 29 '13 at 10:52