Answer:
base: 6 ft by 6 ft
height: 3 ft
Explanation:
The least material will be used when the area of the tank is a minimum.
Area
For base dimensions x and y, and height z, the area of the material used to construct the tank will be ...
A = xy +2xz +2yz . . . . . . . sum of areas of rectangular faces
Setup
There are several ways to choose values of x, y, and z that minimize A. One of them uses Lagrange multipliers. That can be avoided by making the assumption that the minimum area will require the base to be square (x=y). Then we have ...
A = x^2 +4xz
The height will be ...
z = (A -x^2)/(4x)
And the constraint on volume requires ...
xyz = 108 = x^2·z = x^2(A -x^2)/(4x)
Simplifying, we have ...
432 = x(A -x^2)
A = 432/x +x^2 . . . . . an expression for area that we can minimize
Solution
The minimum value of A will be found where its derivative with respect to x is zero:
dA/dx = -432/x^2 +2x = 0
x^3 = 216 . . . . . multiply by x^2/2 and add 216
x = 6 . . . . . . . . cube root
Then the value of z is ...
z = (A -x^2)/(4x) = ((432/x +x^2) -x^2)/(4x) = 432/(4x^2) = 108/36 = 3
Dimensions
The dimensions of the tank that minimize its area for the given volume are ...
base: 6 ft by 6 ft
height: 3 ft
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Additional comment
You may notice that the shape of the tank is that of half a cube. That is, the height is half the length of the side of the square base. This has the effect of making the areas of opposite sides equal to the are of the base. That is, each of the terms in the area expression:
A = xy +2xz +2yz
have equal values: 36.
This is a typical result of this sort of optimization problem. There is nothing about the function for A that favors having the terms be of different values, so the optimum distribution of material makes them all the same.
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If you go through the full 3-dimensional optimization using Lagrange multipliers, you get exactly the same result. The base is square, and each side has half that area.