Expert: Jack Cheng - 11/14/2006

Question
Rectangle ABCD has two vertices on the semicircle y= the square root of 16-x^2 and two vertices on the x-axis. Express the area of the rectangle as a function of the x-coordinate of A. What is the domain of the area function?

Answer
Hi again, Nikki,

It really helps to draw a picture for this question.

The area of a rectangle is l*w, where l is the length (height in this case) and w is the width. So in solving this question, we have to first find the length and width of the rectangle.

Now, let's say the A is the left most vertex on the x axis. The width is easy, because it's just two times the x-coordinate of A (because the graph of the semi-circle is symmetric, the distance from the y-axis is the same).

For the length, it's y-coordinate of the point (B, let's say) on the semi-circle directly above A. In other words, it's the value of the function of the semi-circle evaluated at the x-coordinate of A, which is Sqrt(16-x^2), where Sqrt means the square root of, and x is the x-coordinate of A.

Now that we know both the width and length, the area can be computed, which is 2x*Sqrt(16-x^2).

Now, for the domain, you can do it visually with the graph, or algebraically. Graphically, the largest width of the rectangle is radius of the semi-circle, which is 4. Thus, the x-coordinate of A can anywhere from -4 to 4.

Algebraically, the Sqrt function cannot have a negative input, so 16-x^2 cannot be less than 0. In other words:

16-x^2 >= 0
   16 >= x^2
    4 >= x      (taking square root of both sides)
or  -4 <=  x    

So x should be between -4 and 4, inclusive. That's the same answer we got when doing it graphically.

I hope this helps,
Jack