What are the dimensions of the most economical cylindrical can?

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Homework Help Overview

The problem involves finding the dimensions of a cylindrical can that holds a specific volume while minimizing the cost of materials used for its construction. The subject area includes optimization in calculus and geometry.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the equations related to volume and surface area, questioning the correctness of the original poster's equations and variable definitions. There is mention of using differentiation to find optimal dimensions.

Discussion Status

The discussion is active, with participants exploring different methods for solving the problem. Some guidance on using differentiation to find maxima and minima has been provided, but there is no explicit consensus on the approach or the correctness of the equations presented.

Contextual Notes

There is some confusion regarding variable definitions, specifically the use of "j" instead of "r" for the radius, which may affect the clarity of the equations being discussed.

teng125
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A cylindrical can is to hold 20 m3. The material for the top and bottom costs
€10/m^2 and material for the side costs €8/m^2. Find the radius r and height h of
the most economical can.

the answer is h= 5 and j=2

i have found the eqn such as v=pi j^2 h =20 pi

and 2pi j^2 + pi j h = 18...then i don't know how to do...pls help
 
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Have you learned yet how to use differentiation to find maxima and minima of a function?
 
ya...but is my eqn correct??
 
What's "j"? Where's "r" in the equations?
 
j = radius
 
Well, anyway, I'd solve it this way:

-- write an equation 1 for V = f(r,h)

-- write an equation 2 for A = f(r,h)

-- rearrange equation 1 so you can substitute it into equation 2 to eliminate r in equation 2.

-- differentiate the result with respect to h, and set the result = 0

-- solve for the height h that minimizes the area, and then use equation 1 to solve for r.
 

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