• anonymous
Could please help me ? An open tank is to be constructed with a square base and vertical sides and be able to hold 32 m^3 of liquid when full. Determine the dimensions of the tank if area of sheet metal used to make the tank is to be minimum. :)
  • Stacey Warren - Expert
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  • chestercat
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  • asnaseer
Firstly draw a picture to help you visualise the problem: |dw:1326927864214:dw| I have labelled the side length of the square base 'b' and the height of the tank as 'h'. Now, you are given the volume of this tank is 32 m^3, so we can write this as:\[V=32\]The volume of this tank can also be calculated by multiplying the area of its square base by its height, so:\[V=b^2h\]Since both equations represent the same volume, we can equate them to get:\[b^2h=32\]therefore:\[h=\frac{32}{b^2}\tag{1}\]Next, you are asked to determine the dimensions such that you use the minimum amount of sheet metal. This implies you need to minimize the surface area as this would then use the least amount of sheet metal. The surface area is given by the area of the base plus the areas of each of the four vertical sides. So we get:\[S=b^2+bh+bh+bh+bh=b^2+4bh\tag{2}\]We can now substitute 'h' from equation (1) into equation (2) so that we are left with an expression for S in just one variable 'b' as follows:\[S=b^2+4bh=b^2+4b*\frac{32}{b^2}=b^2+\frac{128}{b}\tag{3}\]So the problem is now reduced to trying to find a value for 'b' that will minimize S. Do you know how to do this?
  • anonymous
  • anonymous
or just go to and search

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