Question

The equation y=sqrtx, x=9, y=0, and a rectangle with its sides parallel to the axes and its left end at x=a. Find the dimensions of the rectangle having the maximum possible area.

Answer 1

Are there any restrictions on \(a\) ?

Answer 2

no

Answer 3

Try and relate the area of the rectangle with the \(\sqrt{x}=y\) function.

You can see that:

So, the area of the rectangle is \(A=(9-a)(y)\) since y=sqrt(x)

But, you know that at the vertical line x=a, you have the point \(y=\sqrt{a}\)

So you get that \(A=(9-a)(\sqrt{a})\)