The cow problem
problem statement
There is a square barn 10 ft. x 10 ft. A cow is attached to a corner of the barn with a rope that is 100 ft. long. What is the area the cow can graze?
Process
When I first saw the problem, what I initially thought was that to help find the area of the circle, we would have to use a compass. Something that I haven't used since freshman year. As we progressed through the problem we realized that all we needed to draw an accurate diagram of the problem was to find out what the radi were from the different corners of the barn.
What helped the most was breaking the whole shape down into smaller shapes so that in the end we add them all up and we have the whole area of the shape.
solution
The equation we used to find the total area of the funky shape was... Area of 3/4 circle + (area of right triangle + area of partial circle/cone)2 - 1/2 the barn. The 3/4 circle has a radius of 100 and after plugging that in to find the area we got 31,400, then we took 75% of that number because we have 75% of the circle and got 23,550. We used the pythagorean theorem to find the length of the line that split the barn in half. We plugged in 10^2 +10^2 = c^2 and got c = 14.14. We divided that in half to get the base of our right triangles and got the base equal to 7.07. We then used the hypotenuse length and base length of the triangle to find the other length using the pythagorean theorem and got the height of the triangle to be 89.7. We then used the inverse sin of 89.7/90 to get the angle measure of the triangle and got 85.3. We added 85.3 and 45 so that we could subtract it from 180 to get the angle measure of the cone shape. We did that and got the cone shapes angle measure as 49.7. We did this so that we could fins how much go the circle we had in order to find the area of that portion, We did 49.7/360 and got .14 which means we have 14% of the circle. We then found the area of the triangle bh/2 and got the triangle's area equal to 317.1. We then proceeded to finding the area of the cone shape which was 3,590. We plugged those numbers back into the equation... 23,550 + (317.1 + 3,590)2 -50 = 30,835.1
maximum rectangle problem
problem statement
A rectangle has one corner on the graph of y=16-x^2, another at the origin, a third on the positive y axis, and the fourth on the positive x axis. If the area of the rectangle is a function of x, what value of x yields the largest area for the rectangle?
process
Before my table group and I had any real guidance from the teacher, we really had no clue where to start. I think a lot of the thoughts that we had were really trying to piece together the bits of information that we had and make connections of rectangles to parabolas. After hitting rock bottom with our observations, we asked for help and finally had something to follow: our equation. The equation allowed us to find what points on the parabola we could use to make a rectangle. Using the equation we found the intercepts and were able to plug in values for x and y. I think our initial attempts were solid and we were on the right path to solving the equation.
solution
This problem is very basic math once you figure out the equations and how to plug in certain things into it. So as soon as we got the equation of what Y was equal too we figured out the Y intercept and the X intercepts after that we plugged in values for the different points for the parabola. After that we tried to draw the different rectangles within the parabola and ended up getting an x and y table that showed us the lengths and widths. After that we calculated the area for those which is just x times y. We saw that 2 had the largest area so decided to go along with that and try all of the numbers between 2 and 3. Instead of just doing x times y, we didn’t know the value of y we had to plug that into the equation so it was area=x(16-x^2) and we would plug in all of the different values of x in the equation. We saw that it started out small and then went up and then fell back down, that is because the equation is for a parabola. The point where it gets the biggest is at 2.3 as our x and 10.71 as our y. Since we found that out we went one step further using the same equation as the one before but plugged in all the numbers in between 2.3 and 2.4. We saw that at 2.31 is the largest that it got because at 2.32 it started to go down in the area.
To find the perimeter of this problem we knew that the equation to find the perimeter is 2(x*y). We first plugged that into the regular x and y table and saw that 0 and 1 had the same perimeter so we would have to go in between them to be able to find the real perimeter. So we decided to go from .1 to .9. Plugging that into our new equation which is 2x+32-2x^2. We find out that .5 gave us the largest perimeter and we even tried to do .51 but that was smaller than 32.50 which was the largest perimeter. x=.5 y=15.75
To find the perimeter of this problem we knew that the equation to find the perimeter is 2(x*y). We first plugged that into the regular x and y table and saw that 0 and 1 had the same perimeter so we would have to go in between them to be able to find the real perimeter. So we decided to go from .1 to .9. Plugging that into our new equation which is 2x+32-2x^2. We find out that .5 gave us the largest perimeter and we even tried to do .51 but that was smaller than 32.50 which was the largest perimeter. x=.5 y=15.75
group test
To prepare for the group test my group and I decided to just change the number and go through the whole thing so that we can see where our group needed help and we figured that we should just go through the whole things. I feel like I really did understand what was going on in this unit and you can see that in how I performed in the individual test. I feel like the way that we practiced made it easier for us to go through and perform well in both of the tests. But I think that the group quiz was just more of a way of teaching the problem then actually everyone working together to solve it.
evaluation
Why is there no