Maximum area limited by circumference 
The circumference is equal to 10, I am to find x and y (as drawn on the picture) so that the area is as great as possible. I have tried back and forth here.
$A = \frac{\pi y^2}{8} +xy$  
$10 = \frac{y\pi}{2} + 2y + 2x $  
When finding a maximum I know derivation is the way to go, but here there is two variables. And this is an introductory course, and derivation with multiple variables is not a part of curriculum. I tried to solve the circumference function for x and y, and then substituting the values into the function for the area. And then finding the derivative for each function separately, but that didn't work. Well, it worked but the book doesn't agree on the answer.  
 A: As AMPerrine points out, the correct formula for the perimeter of the shape is
$$10 = \frac{y\pi}{2} + y + 2x$$
and using it to express one variable in terms of the other (it is slightly nicer to put $x$ in terms of $y$ for this problem) we have that
$$x=\frac{10-\frac{y\pi}{2}-y}{2}=5-y\left(\frac{\pi}{4}+\frac{1}{2}\right),$$
and therefore
$$A = \frac{\pi y^2}{8} +xy= \frac{\pi y^2}{8} +y\left(5-y\left(\frac{\pi}{4}+\frac{1}{2}\right)\right)=-y^2\left(\frac{\pi}{8}+\frac{1}{2}\right)+5y.$$
Now there's only one variable :)
The general idea for this sort of problem is that there aren't really two variables, because $x$ and $y$ cannot be independently chosen - there's some constraint that makes it such that there's only one "actual" variable. In this case the constraint was on the perimeter. Sometimes there is a clear choice of which variable should be the "actual" variable, and sometimes it's up to you, but in any case, you then use the constraint to put one variable in terms of the other.
