Hello Chelsea,
You have made a minor error in part (a), so let's work this problem from the beginning. All linear measures will be in cm.
(a) Express $\theta$ as a function of $r$.
From the information regarding the perimeter of the sector, we may state the following:
$$2r+r\theta=120$$
Now, we wish to solve for $\theta$. We may begin by subtracting through by $2r$ to get:
$$r\theta=120-2r$$
Now, we may divide through by $r$ to get:
$$\theta=\frac{120}{r}-2$$
Thus, we have expressed $\theta$ as a function of $r$.
(b) Express the area of the sector as a function of $r$.
The area $A$ of a circular sector is given by:
$$A=\frac{1}{2}r^2\theta$$
Now we merely need to substitute for $\theta$, and simplify:
$$A=\frac{1}{2}r^2\left(\frac{120}{r}-2 \right)=60r-r^2$$
Thus, we have expressed $A$ as a function of $r$.
(c) For what radius $r$ is the area a maximum?
We could observe that the area function is a quadratic in $r$, with the roots $r=0,\,60$ and thus the axis of symmetry must be the line $r=30$ which is where the vertex is. Since the area function is a parabola opening down, we know the vertex is at the global maximum, and so $A$ is maximized for $r=30$.
You may however be expected to use differential calculus to maximize the function, and so we find by differentiating with respect to $r$ and equating to zero:
$$A'(r)=60-2r=0\,\therefore\,30-r=0\,\therefore\,r=30$$
Now since we have $$A''(r)=-2<0$$, we know by the second derivative test that the extremum we have found is a maximum.
(d) What is the maximum area? in cm^2
To find the maximum area, we let $r=30$ in the area function:
$$A_{\text{max}}=A(30)=60(30)-30^2=30^2(2-1)=900\text{ cm}^2$$
To Chelsea and any other guests viewing this topic, I would invite and encourage you to post other optimization problems in our http://www.mathhelpboards.com/f10/ forum.
Best Regards,
Mark.