# Help Solve Calculus Project - Economical Can Dimensions

• hankjr
In summary, the conversation is about a calculus project involving a pop can. The project has three parts, with the third part concerning finding the dimensions of the most economical can. The can is a 12 oz cylindrical can with no waste involved in cutting the vertical sides, but the top and bottom are made from square pieces of metal which results in wasted material. The task is to find the minimum area of metal needed to make the can, taking into account the excess material from the circular ends. The minimum area can be calculated by minimizing pi*x*h + x^2, with the constraint that pi/8*x^2*h = 12.
hankjr
I was assined a pop can project in calculus class that contained three parts. I did the first 2 parts with little to no trouble but I really need help on the third because I don't even know where to start. Plz help me as soon as possible!

C.) When a 12 oz can is to be manufactured there is no waste involved in cutting the metal that makes the vertical sides of the can ( why?), but the top and bottom are made from square pieces of metal and the corners of the square are wasted. Find the dimensions of the most economical can ( remember that the square corner pieces are not used but would be part of the materials purchased to make the can.

I had some trouble interpreting this, but I think all they are looking for is the minimum area of metal needed to make a 12 oz cylindrical can, subject to the fact that the circular ends of diameter d and raduis r require excess material. The total area of metal would be A = pi*d*h + 2d², with the volume being V = h*pi*r²

hankjr said:
I was assined a pop can project in calculus class that contained three parts. I did the first 2 parts with little to no trouble but I really need help on the third because I don't even know where to start. Plz help me as soon as possible!

C.) When a 12 oz can is to be manufactured there is no waste involved in cutting the metal that makes the vertical sides of the can ( why?), but the top and bottom are made from square pieces of metal and the corners of the square are wasted. Find the dimensions of the most economical can ( remember that the square corner pieces are not used but would be part of the materials purchased to make the can.

There is no waste from the vertical sides because that is a rectangle shaped into a cylinder. Let h be the height of the cylinder, x be the length of the sides of the square from which the top and bottom are cut. I hope it is obvious that x should be the diameter of the the top and bottom (otherwise we could first cut a square to that size and then cut the circle). The radius of the top and bottom is x/2 so the volume of the cylinder is $\pi (x/2)^2 h= \frac{\pi}{8}x^2 h= 12$ (12 oz is a volume.) The waste material will be the area of the square minus the area of the circle: $x^2- \pi/4 x^2= (1- \pi/4)x^2$. You want to minimize that subject to the condition that $\pi (x/2)^2 h= \frac{\pi}{8}x^2 h= 12$.

HallsofIvy said:
There is no waste from the vertical sides because that is a rectangle shaped into a cylinder. Let h be the height of the cylinder, x be the length of the sides of the square from which the top and bottom are cut. I hope it is obvious that x should be the diameter of the the top and bottom (otherwise we could first cut a square to that size and then cut the circle). The radius of the top and bottom is x/2 so the volume of the cylinder is $\pi (x/2)^2 h= \frac{\pi}{8}x^2 h= 12$ (12 oz is a volume.) The waste material will be the area of the square minus the area of the circle: $x^2- \pi/4 x^2= (1- \pi/4)x^2$. You want to minimize that subject to the condition that $\pi (x/2)^2 h= \frac{\pi}{8}x^2 h= 12$.
I think minimizing the waste is not sufficient. x can be made arbitrarily small by making h as big as it needs to be to form a 12 oz can with tiny circles for ends. The total area of metal used to make the can, including the waste, must be minimized.

Ah, you are right. I misread the problem. The problem is to minimize $\pi x h+ x^2$ subject to the condition $\frac{\pi}{8}x^2h= 12$.

## 1. What is the purpose of the "Help Solve Calculus Project - Economical Can Dimensions" project?

The purpose of this project is to use calculus principles to find the dimensions of an economical can that will minimize the cost of materials used to manufacture the can.

## 2. Why is it important to find the optimal dimensions for an economical can?

Finding the optimal dimensions for an economical can can help reduce production costs, which can lead to a more affordable product for consumers. This can also improve the efficiency and sustainability of the manufacturing process.

## 3. What are the key equations and variables used in this project?

The key equations used in this project are the volume of a cylinder (V=πr2h), the surface area of a cylinder (A=2πrh+2πr2), and the cost function (C=ph+2πr2c), where r is the radius, h is the height, p is the cost of materials per unit volume, and c is the cost of materials per unit surface area.

## 4. How does calculus help in finding the optimal dimensions for an economical can?

Calculus helps in finding the optimal dimensions for an economical can by maximizing or minimizing a function (in this case, the cost function) using techniques such as differentiation and optimization.

## 5. What are some real-world applications of this project?

This project has real-world applications in the manufacturing industry, specifically in the production of cans or containers for food and beverages. It can also be applied to other industries where minimizing production costs is important, such as in the production of packaging materials.

Replies
2
Views
5K
Replies
3
Views
3K
Replies
5
Views
962
Replies
13
Views
2K
Replies
14
Views
880
Replies
5
Views
2K
Replies
1
Views
792
Replies
2
Views
3K
Replies
26
Views
2K
Replies
14
Views
3K