How do you find the cost of materials for the cheapest such container given a rectangular storage container with an open top is to have a volume of `10m^3` and the length of its base is twice the width, and the base costs $10 per square meter and material for the sides costs $6 per square meter?

Let `x` and `y` be the lenght and width of the base, and `z` be the height.

Knowing that the lenght is twice the width means that `x=2y`

Knowing that the volume is fixed at `10m^3`, means that `x*y*z=10`

From the previous relation, we can express `x` in terms of `y`, obtaining `x*y*z=10 \rightarrow 2y*y*z=10 \rightarrow 2y^2*z=10`, and from this equality we can obtain `z` as a function of `y`: `z=10/{2y^2}= 5/y^2`

Expressing all the three variables in terms of one is important, because now we have a problem which depends on a single variable: the area of the basis is `x*y=2y^2`, and the lateral surface surface is given by `2(x+y)z=6y*5/y^2=30/y`

Since the base costs `$10` per square meters, and the lateral surface costs `$6` per square meters, the total cost is given by
`10*2y^2 + 6*30/y=20y^2 + 180/y`.

So, `c(y)=20y^2 + 180/y` is the function that we want to minimize, to minimize the cost. Let's study its first derivative, and find the values for which it equals zero:

`c'(y)=40y - 180/y^2=0 \iff 40y = 180/y^2 \iff y^3=9/2`
So, if you choose `y` as the cube root of `9/2`, you'll minimize the function, and thus the cost.