it's not fixed. (you could have checked by expanding what you have here. you'd realize you do not get the original quadratic). there was no problem with the part you changed. the problem was with your constant. i suspect the -3 in front of the y^2 messed you up. remember for completing the square, the coefficient of the squared term must be one.

so you would have to rewrite $\displaystyle A = -3y^2 + 200y$ as $\displaystyle A = -3 \left( y^2 - \frac {200}3y \right)$ and complete the square with what's inside the brackets, then multiply out.

the answer is: $\displaystyle A = -3 \left( y - \frac {100}3\right)^2 + \frac {10000}3$

moving on.

what is the y-value that gives the maximum?