Thread: Finding the cost for the cheapest container?

A rectangular container with no top has a volume of 20 m^3. The length of its base is twice the width.
The cost for the material of the base is $5 per m^2.
The cost for the material of the sides are $12 per m^2.
What is the cost for the cheapest container?

My attempted work:

Let x be the width. Let y = length.
V = 20 = xyz.

C = 5xy + 12(2xz) + 12(2yz)
C = 5xy + 24xz + 24yz

V = 20 = xyz
z = 20/(xy)

C = 5xy + 240/x + 240/y

Cx = 5y - 240/(x^2)
Cy = 5x - 240/(y^2)

Not sure how to get the answer. Can somebody show me how?

Don't forget that the length is twice the width. That is, .

So:







Take the derivative from here.

Originally Posted by drumist

Don't forget that the length is twice the width. That is, .

So:

How did you get

and

Originally Posted by ione

How did you get

and

I copied it from what AlphaRock showed, but it looks like he made a mistake.

We have and , so: