Finding min. value of dimensions for a tank of water

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To find the dimensions of an open tank with a square base that holds 32m³ of water while minimizing the sheet metal area, the volume equation x²h = 32 can be used to express height h in terms of base side x. The surface area, represented by the equation x² + 4xh, can then be rewritten as a function of x alone. To find the minimum area, standard optimization techniques, such as using the second derivative test, should be applied. An alternative approach involves the Lagrange multiplier method for those familiar with it. Ultimately, the optimal design resembles a cube if the tank were closed, suggesting that a square base is efficient for minimizing material use.
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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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