<span> Draw a picture of the downward-facing </span>parabola<span>, and of a </span>rectangle<span> of the type described. Let (</span>x,y<span>) be the </span>upper<span> right-hand corner of the </span>rectangle<span>. Then by symmetry, the </span>base of the rectangle<span> has length </span>2x<span>, and the height is </span>y<span>, that is, </span>12−x2<span>. So the </span>area<span> A(</span>x<span>) of the </span>rectangle<span> is given by A(</span>x)=2x(12−x2<span>).</span>
