right circular cylindrical tin cans are to be manufactured to contain 400 cm ^3 of volume. There is no waste involved in cutting the tin that goes into the vertical sides of the can, but each end piece is to be cut from a square and the corners of the square wasted. Find the ratio of the height to diameter for the most economical cans.
I will assume that by "most economical" you mean
"minimum cost of material for surface area".
let each side of the end square be x, so the radius for the end circles is also x
let the height be y
SA = 2x^2 + 2πxh
but V = 400
400 = πr^2h
h = 400/(πr^2)
SA = 2x^2 + 2πx(400/(πx^2)
= 2x^2 + 800/x
find d(SA)/dx
set it equal to zero and solve for x
straightforward from here.
To find the ratio of height to diameter for the most economical cans, we need to consider the surface area and volume of the tin can.
Let's assume the height of the cylindrical can is 'h' and the diameter of the circular base is 'd'. The radius of the circular base will be half the diameter, so r = d/2.
The surface area of the can (excluding the ends) can be calculated as the lateral surface area of a cylinder, given by A = 2πrh.
The volume of the can can be calculated as the volume of the cylindrical portion, excluding the ends, which is V = πr^2h.
Now, let's consider the ends of the can. Each end piece is cut from a square, and the corners of the square are wasted. Since the can is circular, the square cut out will be in the shape of a circle with a radius equal to the side length of the square. The area of each waste is given by A_waste = πr_waste^2.
To find the total surface area of the can, we need to consider both the lateral surface area (excluding the ends) and the area of the two waste pieces. So, total Surface Area (SA) = 2πrh + 2πr_waste^2.
Since the volume of the can is given as 400 cm^3, we have the equation πr^2h = 400.
Our goal is to minimize the total surface area of the can. In other words, we need to find the height-to-diameter ratio that minimizes SA.
To solve for the most economical can, we can use calculus. We differentiate SA with respect to h and set it equal to zero to find the critical points. Then, we use the second derivative test to confirm that the critical point is indeed a minimum.
Once we have the critical point, we can substitute it back into the equation for the volume to find the corresponding value for the diameter. Then, we can calculate the ratio of height to diameter.
This process requires differentiation and solving an equation, which can be done analytically or using computational software.
By following these steps, the most economical ratio of height to diameter for the tin can can be determined.