Example: An architecture company is designing a new opera house. The basic floor plan is a semicircular area centered around the stage. Due to the location of the building, the final row of seats can only be laid out on an arc measuring 109 feet long. Estimates suggest that for every 3 square feet an additional chair can be placed in the seating area. The owner would like the ground floor to seat at least 2300 occupants with the nearest seats 40 feet away from the stage.
Based on the desires of the owner, constraints of the back row, and estimates regarding how many seats will fit in a given area, what dimensions (radius and angle) of the semicircular area would satisfy the owner? Approximations (rounded answers) are acceptable.
Solution We will need to use both the arc length and the sector area formulas to figure this one out, so let's first jot them down:
\[ s=r\theta \] \[ A=\dfrac{{1}}{{2}}\theta r^2 \]Next, let's sketch out what this will look like:
Based on the arc length, we know that \(109 = r\theta\), so we have a preliminary relation between the two variables we desire to solve for. Next we will need to use the other constraints to break this down even further. Let's begin with the "3 square foot per chair" and 2300 chairs. Putting those two together, we will need a total of \(3\times 2300=6900 ft^2\) to fit the desired number of chairs. However, the area we are looking at is the area 40 feet from the stage, so let's see if we can model that with a formula:
\[ A_{\text{{No Chairs}}} = \dfrac{{1}}{{2}}\theta (40)^2=800\theta \] \[ A_{\text{{Whole}}} = \dfrac{{1}}{{2}}\theta r^2 \] \[ A_{\text{{Seating}}} = \dfrac{{1}}{{2}}\theta r^2 -800\theta \]Since the total area needs to be 6900 square feet, we can rewrite that last equation:
\[ 6900 = \theta\left(\dfrac{{1}}{{2}} r^2 -800\right) \]Of course, we don't want two variables, so now we can go back and use our arc length equation to substitute one for the other:
\[ \solve{ 109 &=& r\theta \\\theta &=&\dfrac{{109}}{{r}} } \]Substituting gives us:
\[ \solve{ 6900 &=& \dfrac{{109}}{{r}}\left(\dfrac{{1}}{{2}} r^2 -800\right)\\ 6900 &=& \dfrac{{109}}{{2}}r - \dfrac{{87200}}{{r}}\\ 6900 r &=& \dfrac{{109}}{{2}}r^2 - 87200\\ \dfrac{{109}}{{2}}r^2 - 6900r -87200 &=&0 } \]Now, while that looks like: "ugh, really?" we can use the quadratic formula or a graphing utility to get an approximate decimal answer:
Since radius is a distance, we will drop the negative answer and accept the positive answer of 138 feet. Finally, the angle we can quickly determine (in radians) from that arc length relationship:
\[ \solve{ 109 &=& r\theta\\ 109 &=& 138\theta\\ \frac{{109}}{{138}}&=&\theta\\ 0.79&\approx &\theta \\45.3^\circ&\approx&\theta } \]