# BY Ashlynd broling, michael volkening, and Rachel holesha

# Linking Cans, Cylinders, and Rational Equations

For this project/presentation, we were required to solve a set of problems revolving around the dimensions of a cylinder and rational equations. Here you will see our work, answers, and our explanation on how we came to that solution.

These pictures are our work for the four problems we were asked to solve. All our work is shown and a short explanation of how we got what we did is given in these pictures.

# Problem Number 1

In this problem we were asked to determine the formula for the volume of a can in terms of h, r, and pi. We did this by using the basic shapes that form to create a can. These shapes where one rectangle and two circles. We were then able to see where the terms h, r, and pi come into play. The h came from the height of the rectangle and the r came from the radius of the two circles. We then came to the conclusion that by combining the shapes and terms of those shapes we could come up with the volume of a cylinder formula. The formula we came up with was V=pi(r^2)(h).

# Information on Finding the Volume of a Cylinder

https://www.youtube.com/watch?v=y3TjAHV7esk from Vividmaths.com

# Problem Number Two

In this problem we were asked to determine a formula for the cost of a coffee can based upon a diagram that we made. We determined that a formula for the cost would just be the area of the two circles of the can and the area of the part of the can that made a rectangle. So what we did was combined the area of the two circles and the area of the rectangle by adding them. So in that case the cost of a coffee can would be equal to the area of the two tops of the can (two circles) plus the area of the outer edges of the can (the rectangle). (Cost= 2pi(r^2) + 2pi (r)(h)).

# Problem Number Three

In this problem we were asked to find the specific cost of the coffee can mentioned at the beginning paragraph of the project. The specific requirements that have to be used for this can are an area of 135 in^3 of coffee grounds. Six cents per square inch of thick aluminum on top of the can and the bottom. Then finally thinner grade aluminum that is four cents per square inch on the sides of the can. The equation that we ended up coming up with for the cost of this can was 0.12pi(r^2) + 0.08(135/r). We obtained this equation by first taking the equation Cost=pi(r^2)(h) + 2pi (r^2)+ 2pi(r)(h) then we solved for h by using 135=pi(r^2)(h) then we plugged h into the starting formula and wee got the end formula. Also we plugged in the needed values of the can into the equation. (As seen the 2(0.06) pi (r^2) + 2(0.04) pi (r)(h) equation originated from the Cost= pi(r^2)(h) +2pi (r^2) + 2pi(r)(h) it is just after solving for h and plugging it in.

# Problem Number Four

In this problem we were asked to find the minimum cost of the coffee can, the dimensions of the coffee can, and then to explain how we found this information. The way we found the minimum cost of the coffee can was we simplified the equation 0.12pi (r^2) +0.08(135/r) and we found a common denominator for each side of the equation. It turned out to be ((0.12 pi(r^2) + 10.8))/r. We then plugged this equation into the calculator and used the second calc from and clicked minimum which gave us a right left bound graph. We used the right bound portion because the cost must be positive and went to the lowest portion on this part of the graph and found the minimum cost to be $6.68. We then used the simplified cost equation 6.68=0.12pi (r^2)+0.08(135/r) and solved for r. This gave us the radius of the can. Then we used the volume equation put to h to solve for h (h=135/pi(2.34)^2). (2.34=r) These found dimensions would make sense because the height is logistically proportional to the radius. Also when you plug it into the equation we plugged into the calculator equal to itself it works out.

# Finding the minimum value of the equation

http://resources.chuh.org/staff/aklein/Alg2/TI_max%20min%20zeros.pdf from TI Calculator Resources

# The cost of a pop can?

Using our own pop can we are now going to find the minimum cost of it and the dimensions. Here is the picture of our can with the found dimensions of a radius of 1 inch and a height of 4.75 inches.

# Minimum Cost of our Can

We found that finding the minimum cost of a can with already having the volume, height, and radius is a lot easier than without having it just like the problem we had to do. In order to find the minimum cost of the coffee all that is needed is to add the cost of the volume of the can, the cost of the two circles, and the cost of the outside of the can. However, in order to be more accurate the formula can still be used however is is tweaked a little bit from the original one from the other problem. What we did was take the formula 1.2pi r^2+ 0.8pi (14.92/r). This equation was originated from the last minimum equation from the previous problem. Then you take this and plug it into the calculator and find the minimum just like it was done in problem number four. It was found that the minimum cost of the coke can was 15.37. This number is seemingly extremely off because of the volume of the can. This throws off the total cost of the can. The height of our can was 5 inches and the radius was 1.25 inches.