Finding min. value of dimensions for a tank of water

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Homework Help Overview

The problem involves determining the dimensions of an open tank with a square base that can hold 32m³ of water, while minimizing the area of the sheet metal used for its construction.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the relationship between the volume and surface area of the tank, with one suggesting the need to express height as a function of the base side length. Others mention using techniques such as the second derivative test and Lagrange multipliers to find minimum values.

Discussion Status

Several participants are exploring different methods to approach the problem, including hints about substituting variables and considering alternative optimization techniques. There is no explicit consensus on a single method, but productive suggestions have been made.

Contextual Notes

One participant notes their background in a physics foundation year, indicating potential gaps in knowledge regarding certain mathematical methods mentioned in the discussion.

BOAS
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Hello

Homework Statement



An open tank is constructed with a square base and vertical sides to hold 32m3 of water. Find dimensions of the tank if the area of the sheet metal used to make it is to have a minimum value.

Homework Equations





The Attempt at a Solution



I'm not entirely sure of how to approach this problem beyond needing to use the second derivative. I think I need to construct an expression of the area related to volume.

So,

the sides pf the base, x multiply together to give an area x2 and the four sides can be called 4xh (side of the base multiplied by height)

x2 + 4xh is an area of sheet metal

x2h = 32

I don't know how to proceed.

Thank you for any help you can offer
 
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BOAS said:
Hello

Homework Statement



An open tank is constructed with a square base and vertical sides to hold 32m3 of water. Find dimensions of the tank if the area of the sheet metal used to make it is to have a minimum value.

Homework Equations





The Attempt at a Solution



I'm not entirely sure of how to approach this problem beyond needing to use the second derivative. I think I need to construct an expression of the area related to volume.

So,

the sides pf the base, x multiply together to give an area x2 and the four sides can be called 4xh (side of the base multiplied by height)

x2 + 4xh is an area of sheet metal

x2h = 32

I don't know how to proceed.

Thank you for any help you can offer

The only thing you're missing is that your volume function can be solved for h as a function of x, and then substituted into your surface area function, making it a function of x alone. Once you have the surface area as a function of x, use the usual technique for finding the minimum value.
 
Mark44 said:
The only thing you're missing is that your volume function can be solved for h as a function of x, and then substituted into your surface area function, making it a function of x alone. Once you have the surface area as a function of x, use the usual technique for finding the minimum value.

ah of course! Thanks
 
BOAS said:
Hello

Homework Statement



An open tank is constructed with a square base and vertical sides to hold 32m3 of water. Find dimensions of the tank if the area of the sheet metal used to make it is to have a minimum value.

Homework Equations





The Attempt at a Solution



I'm not entirely sure of how to approach this problem beyond needing to use the second derivative. I think I need to construct an expression of the area related to volume.

So,

the sides pf the base, x multiply together to give an area x2 and the four sides can be called 4xh (side of the base multiplied by height)

x2 + 4xh is an area of sheet metal

x2h = 32

I don't know how to proceed.

Thank you for any help you can offer

Besides the hint already given, you can also use the Lagrange multiplier method (if you know about it).
 
Ray Vickson said:
Besides the hint already given, you can also use the Lagrange multiplier method (if you know about it).

I don't know about that.

I'm in a physics foundation year (year 0 at my uni) due to not having done the traditional subjects that lead to taking a physics degree.
 
There is a shortcut available here. If the tank also had a top, you'd straight away say it should be a cube. With the open tank, imagine creating a closed tank by taking two optimised open tanks and inverting one over the other. You must now have created an optimal closed tank, i.e. a cube.
 

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