Optimization Example 4

 I've been sent another maximization or minimization problem, and this seemed to be an interesting twist on what we've already done, so let's do this one. Let's see. A rectangular storage container with an open top is to have a volume of 10 meters cubed. The length of its base is twice its width. Let me start drawing this thing. It's a rectangular storage container with an open top. So let's see, let me do it a different color.  OK, so I'll do it over here. So that's its base.  So let's see. The length of its base is twice its width. So this is the base, and if I call this the width, so this is x, the length is twice that, this is 2x. The length of its base is twice the width, right? So this is the length of the base, this is the width. OK? Material and then the height, for now, is unknown. I think we should be able to find a constraint on this, because we know what the volume has to be, right? It says, materials for the sides of cost $6 per square meter. Find the cost of materials for the cheapest such container. OK, so let's do a couple things. First, let's figure out what the height is in terms of x, and then we can find an equation for the cost in terms of x, and then take the derivatives, and find the minimum, et cetera, et cetera. So what do we know about this? What's the volume of this box? The volume of this box is equal to the, what did I say, this is the length, but it doesn't really matter. Length times the width times the height. So it's 2x times x, 2x squared times h. So it's 2x squared times h, right? Base times height times width, whatever. Length times width times height. Just the volume of a rectangle. And they already said that the volume is 10 meters cubed, so that gives us a good constraint. So that has to equal 10. Let's divide both sides by 2x squared, and so what do we get? If...