I’ve been sent another maximization or minimization problem and this seem to be an interesting twist on what we have already done so let’s do this one. You 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. So let me start drawing these things. A rectangular storage container with the open top, so let’s see, let me do it in a different color. I’m doing this color. So I'll do it over here, so that its base. Let’s see.
The length of its base is twice its width so if this is the base and if I called this the width; this is X. the length is twice that. This is two X. The length of its base is twice the width. So this is the length and the base. This is the width. A material and then height for now is unknown. I think we should be able to find the constraint on this because we know what the volume has to be right? It says material for the sides cost $6.00 per square meter. Find the cost of materials for the cheapest such container. Okay, so let’s do a couple of 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 I take the derivatives and find the minimum etcetera, etcetera. So what do we know about this? What’s the volume of this box? The volume of this box is equal to the—when I say that this is the length but it doesn’t matter in length times the width times the height. So its two X times X, two X squared times H. So it is 2X2 times H. Base times height times width and whatever. Length times width, time height. It’s just the volume of a container—of a rectangle.
And there he said that the volume is 10 meters cube so that gives you good constraint so that has to equal 10. Let’s divide both sides by two X squared and so what do we get? If we are to take this equation, we divide both sides by 2X2 then we get H is equal to 10 over two X squared which is equal to—we could say five X to the minus two power so this is. We could write this as 5X-2. That is H. we have just solve that, and we were able to solve that because they gave us the constraint on the volume of the box. Fair enough.
Let’s see what the cost of the material, the total cost of the material is going to be. I’ll do that in magenta. The cost of the material and I could even say as the function of X is equal to what? Material for the side cost $6.00 per square meter so what is the surface area of the sides? I don’t know, they tell us the material for the base first. Material for the base cost $10.00 per square meter. So this base right here, the material cost $10.00 per square meter. What is the cost of the base? It’s going to be area of the base times 10. So it’s 10 times the area of the base. What’s the area of the base? It is two X times X.
So it is two X squared. So this is the cost of the material o the base and now we want to add the cost of the material on the sides. This is where we get a little bit interesting. So there’s two sides that are X. That this side and this side are the same size. So what are they? What are their areas? It’s going to be $6.00 per square meter for those two sides. For this green side and that one back there as well. I don’t want to make this too messy. So $6 dollars and what’s the area? And if there are two of these sides, I'm going to multiply it by two and so what’s the area of each of these sides? It is X times five X minus two so that is five X minus one.
Now, we have to figure out this side here and that side over there. I will do it in a brown maybe. Now we care about this side up in front and then that side back there but that’s also when it cost $6.00. So plus $6.00 per square meter and then there’s two sides and then what is it? Its two X times five X to the negative two so that’s ten. Two times five is ten. X times the negative two, X to the negative one.
Let’s simplify this. We have the cost, and there is no roof instead it was an open top so there is no fabric we need for the top. So the cost of putting the fabric or the material on this open rectangle is 20X squared plus. What’s this 12 times five is 60X to the minus one plus—let’s see, twelve times ten is 120X to the minus one. And so we can further simplify that as 20X2+ 180X-1. And now we are ready to just take the derivative and find out what the minimum point is, of this curve. So let’s do that.
It’s C prime of X—and this is a fun non-hairy derivative so it is 40 then two times 20X and then the minus 180X-2. The minimum or maximum point are going to be where this is equal to zero so let’s set that equals zero. Let me clear up some space.
I think you understand all of these so let me get rid of that. Actually, I like the drawing but it has got to go. I need the room. If you’ll review it, just rewind the video. We want to set the derivative to find the minimum or maximum point and minimum or maximum point of the curve and the slope is zero. So the derivative is zero so let’s figure that out.
40X-180X-2 is equal to zero. The first thing I would do is just we could divide both sides by 20 and you get 2X-9X and minus two is equal two zero divided 20 is just zero. Then let’s see what you get. If you add 9X to the minus two to both sides you get 2X is equal to 9X-2 or nine over X squared. That’s the same thing as nine X to the minus two.
Now I don’t know. Let’s multiply both sides of this equation by X squared and so you get 2X3 is equal to nine over X to the third is equal to nine halves or X is equal to the cube root or we could just say nine halves to the one third power whatever that could be is four point five to the one third power. Do we know that this is the minimum or maximum point? We could go and take the second derivative and confirm that were concave upwards at this point and then we would know that we are at a minimum point and this is the cheapest such container.
If you see this on and maybe you want to prove that for yourself on a test but an easy way to test that is just try some other X. we have the cost function right here. Try out some other X. You know that this is either going to be a maximum or minimum or at least the local maximum or minimum. If you try another X and if it’s greater than you know it’s minimum. Or the other thing is, when you know if there’s only one critical point and they ask you to find the cheapest container that critical point is probably going to be the minimum point.
Anyway, just for fun let’s figure out what that is actually equal to. I will actually go to Google. Google.com and let’s see. We could just say what we want, nine halves of the one third power. So that is four point five to the one divided by the third power. One point six five so that’s our answer. If X is equal to 1.65+ some other points, then we have the cheapest container to produce. It is 1.65, something, something and there you have it.
Hopefully, you found that very useful.
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