1. Consider a semicircle with radius 1, horizontal diameter PQ, and tangent lines at P and Q. At what height above the diameter should the horizontal line be placed so as to minimize the shaded area?

2. Consider a region consisting of all points inside a square that are closer to the center than to the sides of the square. Find the area of the region.

2. Originally Posted by Vonhohenheim150
1. Consider a semicircle with radius 1, horizontal diameter PQ, and tangent lines at P and Q. At what height above the diameter should the horizontal line be placed so as to minimize the shaded area?

2. Consider a region consisting of all points inside a square that are closer to the center than to the sides of the square. Find the area of the region.

I'm almost sure I know what's that "shaded area" in problem 1, but I won't even bother trying to write down a hint since I can't be sure. I think posters should be more careful when redacting their questions and, to say the least, shouldn't assume people reading their post are prophets or guessers.

As for two: what have you done so far? Do you already know at least what's the shape of all those points?

Tonio

3. Picture to second problem is here:
http://www.stewartcalculus.com/data/..._cp_05_stu.pdf

It's problem 16 there.

4. For question two I believe you need to examine a triangular section of the given shape that only shows one of the four sides. Find the area of the section that consists of points within the triangle that are closer to the center than to the side of the square, it is a triangle plus a crescent-shaped area, then multiply this by four to have the area of the entire shape. Do you understand the basic idea of this? I am not the best at explaining things clearly I'm afraid.

5. I finished the one about the semicircle myself using a different method involving the creation of a ray from the origin towards the point of intersection between the movable horizontal line and the semicircle. Then I defined the shaded areas in terms of that angle and t (where t = the height of the horizontal line). Anyways, I got the (sqrt3)/2.

Still need help with the problem that I had a picture for though : (

6. Squareroot three over two? How exactly did you find the area using that angle?

7. I drew a ray from to origin towards the point of intersection (between the semicircle and the line y=t, where t is the height of the line above the x axis). This would create 2 sectors.
The first sector would have an area of (Omega/2pi) X (pi) which means (Omega/2). Now, since I only want Area 1 and not the entire sector, I subtract the area of the triangle I showed from the area of Sector 1. This would give me Area 1.

Next I need to find Area 2. I do the same thing as last time, except this time I find the area of Sector 2. This is equal to (90-Omega)/(2pi) X (pi) to be (90-Omega)/2. However, this includes the area I sprayed in red. So, I just take the area of Sector 2 and subtract the area of the same triangle from last time (that's because both triangles are the same since they form a rectangle). This gives me the area I sprayed in red. Now I find the area of the box I bordered in green. This is found by multiplying base times height which is:
(1-sqrt(1-t^2)) times t
I take that resulting area of the green box and subtract the area sprayed in red to finally get Area 2.

I add the formulas for both areas to get a single formula representing Total Shaded area. Take that an minimize it by setting the derivative equal to zero and solving and you would get (sqrt3)/2