First I'd let the width of the rectangle be 2y so that the radius of the semi-circle is y.

Then 2x + 2y + pi y = 300 and you can re-arrange this to make x the subject.

A = 2xy + pi y^2 and you can sub x in terms of y from the above.

So you have A as a function of y.

Now solve dA/dy = 0 for y and test that this corresponds to a maximum.