What are the dimensions of the most economical cylindrical can?

In summary, the conversation discusses finding the most economical dimensions for a cylindrical can that can hold 20 m3. The top and bottom of the can cost €10/m^2 and the side costs €8/m^2. The equations for volume and surface area are provided and the conversation mentions using differentiation to find the minimum surface area. The final solution is h=5 and r=2.
  • #1
teng125
416
0
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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  • #2
Have you learned yet how to use differentiation to find maxima and minima of a function?
 
  • #3
ya...but is my eqn correct??
 
  • #4
What's "j"? Where's "r" in the equations?
 
  • #5
j = radius
 
  • #6
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.
 

What is an "Extrema physics problem"?

An Extrema physics problem involves finding the maximum or minimum value of a physical quantity, such as velocity, acceleration, or energy, under certain conditions or constraints.

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Some common types of Extrema physics problems include optimization problems (finding the maximum or minimum value of a quantity), related rates problems (finding the rate of change of a quantity), and constrained optimization problems (finding the maximum or minimum value of a quantity subject to certain constraints).

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To solve an Extrema physics problem, you typically start by identifying the physical quantity that needs to be optimized, and then using mathematical techniques such as differentiation, integration, and setting up equations based on the given conditions or constraints. You then solve these equations to find the maximum or minimum value of the quantity.

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Extrema physics problems are important because they allow us to understand and predict the behavior of physical systems. By finding the maximum or minimum values of physical quantities, we can determine the optimal conditions for various processes and design efficient systems.

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Extrema physics problems have many real-world applications, such as in engineering, economics, and biology. For example, they can be used to optimize the design of structures, determine the most efficient way to produce goods, or find the optimal conditions for biological processes.

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