Another common optimization problem is to determine the dimensions of a box so that the volume is maximized, given the surface area of some material. An open box can be formed by cutting out a square from each corner and folding the sides up--the goal of this problem is to find how large the squares should be in order to maximize the volume.
In this example, let the material be a square so that all sides are the same length--call it L. If the side of the square we want to cut out is x, the side length of the material after we cut out squares from each corner is L-2x. This will be the side length of the base of the box.
Remember that the volume of a box is the surface area base multiplied by the height. We know that the base surface area is A=(L-2x)2. The height of the box will be x, the side length of the square that are cut out. So, the volume of the box will be V=(L-2x)2x. You can find the value of x that will give the maximum volume by finding the maximum of this formula for the volume (recall that this is done by taking the derivative).
Let’s do another optimization problem. Here’s another classic; the box problem. A square sheet of cardboard 67 meters on one side will be used to construct an open top box, by cutting a square from each corner, and folding up the sides.
So here’s what we are starting with. This is our square piece of cardboard, we are going to cut these little tiny squares out of each corner, and then fold up the sides. We will get a box looking like this open on the top. We want to find to find out how large a square should be cut in order to maximize the volume.
So this is a pretty complicated optimization problem, and we will be able to use the closed interval method on this one.
First of all we need to define some variables. I’m going to make x the size of the square that we are going to cut out. So this will be x, this will be x. It's an x by x square, all these are x by x squares. And we will have to also keep in mind that this is a 60 cm square, so this total length is going to be 60. Same thing with this length, 60.
The important dimensions are going to be the dimensions of the base, because the volume is going to be the area of the base. You can kind of see the base, the square base, times the height. The height after you cut up these squares and fold up is actually going to be x. Can you see that? This blue length here when you fold it up, it's going to be exactly the height of the box.
The question is what’s this dimension’s going to be? Since it's going to be square based box, this will be the same dimension but I have to kind of figure out what length that’s going to be. I’m starting with 60cm on each side and I’m cutting out x and x, 2x. So this is going to be 60 minus 2x. So I’m looking at a volume function v(x) equals (60 minus 2x)². That’s the square base times the height x. So it’s the area of the base times the height. That’s the volume of this box.
Now I’m going to want to expand this, it will be easier to work with if I expand it. So let me do that up here. V(x) equals, and then my 60 minus 2x squared will be 3600 minus, and then 2x times 60 is 120x. And I have to double that, 240x. So this is just going to be minus 240x and then I’m going to have to square -2x. So I'll get plus 4x², all of that times x.
So one more step. 3600x minus 240x² plus 4x cubed. I want to use the closed interval method on this problem. Remember, I’m trying to find the maximum volume, but I need a closed boundered interval over which to search for that maximum volume. So I need to look back at my problem here and ask the question what values of x make sense?
First of all I need the minimum value. It’s pretty easy to see that I have a minimum value of 0. If x is 0, I will get a volume of 0 because I will be cutting out squares of 0 size out of each corner. So I’m going to get 0 volume. So I definitely want x to be bigger than or equal to 0, but what’s the maximum? If you imagine how big x could possibly be, I could actually just cut 4 squares out of this thing. That’s the biggest it could be.
Square that's actually this big. That’s going to be on one side, half of 60, so 30. So there is my closed boundered interval. You need this because you need endpoints to use on your table, when you get the table of values using the endpoints and critical points.
So I’m going to write that again up here. X is between 0 and 30, and now I’m going to need to find the derivative. V'(x) is going to be 3600 minus, the derivative of this term is 480x, plus and I’ll get 12 x squared and hopefully this will factor. It looks like it has a factor of 12. So I’m going to take that out and that leaves 300 minus 40x plus x². This looks like a tough factoring problem here. I’ve got a 40x in the middle and I got a 300. I have to think about factors of 300 that add up to 40.
I’m thinking 10 and 30 will work. So we have 30 minus x, 10 minus x. Let me just check that. So 30 times 10 is 300, that works. Minus 10 x minus 30x is -40x, that works. So this is my factored derivative. And the reason I factored, is I want to be able to identify the critical points. And now I can see that the critical points are x equals 10, and x equals 30. Those are my critical points. And now I’m ready to make a table of values; X and v(x ). I want to include the endpoints; 0 and 30.Those are the two endpoints and I want to include the critical points. I put too many slots here I was thinking about two critical points, but one of the critical points is the endpoint. So I just need 10.
Let’s plug these numbers into our volume function. I think the easiest version to plug into is the factored version. Let’s go over here. So when I plug in 0, you can see when x is 0, this whole volume is going to be 0. So that’s nice and easy.
Let’s do the other endpoint, it will probably also be easy. 30, when you plug in 30 you have 2 times 30 which is 60, 60 minus 60 is 0. So this is going to be 0. And then x equal 10, 60 minus 20 is 40. 40² times 10 that’s going to be my volume. This is going to be 1600 times 10, 16,000. So that’s your maximum volume.
Let’s just go back to the problem see what it asks me. How large a square should be cut in order to maximize volume? What I'm really being asked here, is the size of the square that should be cut out, not what the actually maximum volume is.
Although I get that for free. So this is the maximum volume, it’s going to be in cubic cm. But what I really needed to know was, how big a square to cut out. I should cut out a 10 by 10 square. Or rather four 10 by 10 squares, 10cm squares to maximize volume. And that’s our solution, using the closed interval method.
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