Problem Statement
A rectangle has one corner of 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
Since I wasn't present during the week when this problem was introduced I did not have any attempts in solving this problem. However it was introduced to me and explained and I was able to understand how to solve it
SolutionIn solving this you have to find certain answers to formulas like the area and perimeter. First we have to draw the parabola and we had the formula for it and it was y=16-x^2. Given that, we know that the y-intercept is 16. To find the x-intercepts, we plug in for x like 4, 16-4^2 and the x-intercepts were (4,0) We then created a t-table showing the points
x= 1, 2, 3, 4 y=15, 12, 7, 0 |
SolutionArea:
Once we finished with placing the parabola we then try to look for the maximum area by creating rectangles in the parabola and attempting to find the area, the formula being a=x*y. We plug in y for the formula that was provided to us in the beginning. Then we plug in x for any number. When we find the biggest area we start going in by tenths in decimals and the biggest area that we got was: x=2.31 y=10.663 Area=24.63153 |
Perimeter:
Once we are finished with finding the area we move on to finding the perimeter with the formula being p=2(x*y). Similar to finding the area, you plug in y for the formula that was provided in the beginning. Plug in any number for x and the main focus is finding the biggest perimeter.
x Perimeter
0 32
.5 32.5
1 32
2 28
Once we are finished with finding the area we move on to finding the perimeter with the formula being p=2(x*y). Similar to finding the area, you plug in y for the formula that was provided in the beginning. Plug in any number for x and the main focus is finding the biggest perimeter.
x Perimeter
0 32
.5 32.5
1 32
2 28
Group Test/Individual TestIn preparing for the group test we tried getting everyone in the table involved and staying on the same page. We were given a similar question as the original but just switched the number to a different one. We all solved for the area, perimeter and lengths and widths. If anyone in the table got confused or wanted clarification we stopped and tried to explain it to them until they understood.
When we took the group quiz we got stuck on the part where we had to find the area because we had to change the formula of the area from a=x*y to a=(2*x)y because we had to double the x and leave the y because it doesn't get lower or higher. We had a difficult time with this problem because we spent time in looking for a formula that works. We all worked together in trying to figure it out and at the end we did find a formula. I think that our group worked well in communicating in the group quiz because we were explaining to one another what we got or how we were solving it. In the individual it was easier because we just had to plug in for x in the formula like perimeter or area and that was easy because it assisted us in finding the answer. |
Evaluation/ReflectionIn this problem, what pushed my thinking was the group quiz because we had to change the formula of the area an that challenged me and my table and we were working together to find one. If I were to grade myself on this unit I think I would get an A because I looked forward to learning math and I think that I learned more this unit than freshman and sophomore year combined. In this year I am pushing myself to accept the challenges in math so I can improve and so far I have been improving.
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