# Thread: Lost in this one, dimensions

1. ## Lost in this one, dimensions

Find, correct to five decimal places, the dimensions of the rectangle of largest area that has its base on the x-axis and its other two vertices above the x-axis and lying on the curve $y = cos x$ .

I have no idea where to begin... I don't even be clear what it wants for the result

A = bh = x.cos(x)

and then evaluate A'=0 to find the points but I got stuck in the process, I missed something or maybe I'm evaluating the wrong area.

3. Originally Posted by dax918

A = bh = x.cos(x)

and then evaluate A'=0 to find the points but I got stuck in the process, I missed something or maybe I'm evaluating the wrong area.
If "x" is the length of the base, as well as the x-coordinate of a point, then you must have the other vertex as x=0. But their y= cos(0)= 1 so, unless cos(x)= 1, you will not have a rectangle.

But this problem, as stated, has no solution. Taking the left side at $x_0$, we can take the right side at $x+ 2n\pi$ having arbitrarily large area. There is NO "largest area".