Buzz World 365 Logo

Ex: Optimizing the Volume of a Box With Fixed Surface Area

Aug 23, 2022

Ex: Optimizing the Volume of a Box With Fixed Surface Area

Imagine you are in a real application. Most of what you try to do in real life is optimize for different things that you want to minimize the energy in your system. You want to minimize the heat in your system. Whatever it is, it's either a maximization problem or a minimization problem. This is one of the reasons why everyone needs to learn math and STEM disciplines, because things need tweaking everywhere, so in this particular section of our calculus course we're going to have a few different examples where we'll find a method to do the Using tools that we've already developed to apply them to different optimization problems now in this first optimization problem I'm looking at I'm going to talk about a box it's a square bottomed box so the bottom is a perfect square, then it goes up a bit and it doesn't have a top but it has a bottom, now the first step of most optimization problems is to draw a picture.
ex optimizing the volume of a box with fixed surface area
We want to write down all the labels for our variables and just have a very clear picture of what's going on. So let's draw a box there. We have my kind of horrible diagram that's fine if you need it. To have a clearer picture here I'll let my helper show you what this box looks like. Okay James let's teach our students some calculus Height I know let's spin around so in this case I have a square box and what I mean by a square base is this bottom and this side are the same , that's what it means to be a square so let's call that maybe I'll call that bottom length L and there's actually a lot of ELLs there's an L there and there and there and they're somewhere along the bottom will be an L and they are there too they are upstairs there are elves all over the place but let's put the one down and then we'll do the height of it too.
ex optimizing the volume of a box with fixed surface area
I'm going to call that H, so I have a length and I have a height in all these different places. Now I want to go and try to write down some equations that reflect this peculiarity of the box. Well, the first equation that I'm going to write down that relates to this box relates to the constraint. The limitation with this problem is that I only have 12,000 square inches of material to work with out of the 12,000 square inches of material that I need to make. The base I need to make the four different sides so the total

area

here is constrained by those 12,000 square inches will, so this is the formula I get.
ex optimizing the volume of a box with fixed surface area
I call it my constraint equation and what makes sense is I have my

area

my constraint my 12,000 and then this is equal to L squared which is L squared which is the base of my particular box, it has a here L and then also an L there so the

area

of ​​this square on the floor is an L squared I also have this thing called or LH Y for lunch Well let me say this panel right here has a length of L at the bottom and then a height of H, so that rectangle right there will have an

area

of ​​L by H, but I have four of these sides, they're all identical, so four by L by H is my constraint.
ex optimizing the volume of a box with fixed surface area
The next equation I want to write down I call my optimization equation because the thing I want to optimize I want the thing to be as big as possible like mine The

volume

is the blue so they have another formula for my optimization equation and here it is the

volume

of l squared H. This is the

volume

of a box where both the base is the same width and length go to L and then the height H, so L squared H. Now you know we that I want to optimize a specific function of a variable. You take the derivative and set it equal to zero, but the problem is I can't just take the derivative directly of this

volume

and set it equal to zero because in terms of what there are several different variables, here's an L and here an H, if Folia was a function of just one variable, then I could use our method that we saw earlier and set its derivative equal to zero, so I want to take this

volume

, it has two variables and I want it in relation write to just one variable, how can I do that?
ex optimizing the volume of a box with fixed surface area
My trick is that I take the constraining equation. I'm going to try and feed them into my optimization equation I want to get rid of one of the variables I need to get rid of, specifically looking at this I can get rid of L or H but it only misses once and the L comes at two different places so it makes sense to me let's make H blah right now and we'll take the H and put it there then I'm just left with LS. I'll show you how. So first let's take a constraint equation let's rewrite it I solved this for H I got the 1200 across.
ex optimizing the volume of a box with fixed surface area
Rearranged there is my formula for H. I'm going to take the H now. I'm going to put the hat down in the

volume

and where there used to be an alter is now this big messy thing, okay, so I've got this formula, let's bring this up, okay, I've got that, it looks pretty good, but I gonna clean it up a bit so i'm gonna go and just take this L squared here and cancel out the L which just makes an L and then the L squared over L is 1 L times an L squared which makes my L cute makes it a little cleaner alright I'm making progress because I can differentiate now.
I can differentiate because it's just a function of some variable l, so that's pretty good, so let's take the derivative of that, that's just a polynomial, and now we can do that, so its derivative is the derivative of the

volume

with reference to the length is ok 300 the L goes to a 1L cube goes to 3L squared and what I do is I set this equal to zero. I want to find the critical number I want to find out where a candidate is a maximum or a minimum to be asst ok a relatively simple equation this is just a square I can take the 3 l squared move it to the other side , which takes three thousand multiplied by 4/3 square root and what I get, I get that L is 20 root 10 centimeters, by the way, because technically the square is plus or minus 20 root 10 centimetres, but let's think about that the length using a negative number to construct a block doesn't make any physical sense, so I'm just taking the positive square root, so this is my critical number and that's the value of L where I'll have a maximum, or at least I'm completely confident that I mean , it's a critical number for sure, its derivative is zero, but if there's a maximum, there might be a minimum, there's a couple of things I'd like to check first, that the endpoints considered that none of them physically make a box with

volume

as I certainly can't have negative values ​​of L which doesn't make sense, a negative length doesn't make sense but if L is zero then this is a box with Nr base at all that has no

volume

if L is infinity this is like an infinite plane on the ground which would force my height to zero so again it would have no

volume

so the endpoints don't make sense now I can do that reason Since the maximum must occur at either an endpoint or a critical number, and these are clearly places where this has positive

volume

, we can box the positive number at which the maximum occurs at that critical number must as it doesn't occur on the other endpoints As far as I can think this through are the first primitive tests.
So actually looking at what this derivative is, if I make my L a larger number because of this minus sign here, if L is greater, if my L is less than this certain critical number, then my derivative becomes negative, then I subtract less and so I have a positive number so what happens to my function, it goes from increasing to decreasing. I have a maximum over the first derivation test, so it all means that the critical number we calculate is really the maximum now we have this l value this l is 20 root 10 centimeters but what did I ask for the task I asked which one dimensions give the maximum

volume

so i also had to figure out the li the age so let me just address this l here and then i will make an addition that we had a formula for what the formula for h was that it was exactly this thing that we previously said H in relation to L so if I take this particular value of L and drag that along for my age then I now have a new value of H it's 10 root 10 centimeters end point .
Let's think about how this actually looks physically like I tried to draw it badly but at least relatively accurately to what would be the maximum

volume

you can see how the length here is 20 root 10 and the height 10 root is 10, so the maximum

volume

of this open-topped box has a height half that length okay now let's just summarize the main steps we took in each optimization A problem is actually a little different from the others, but there is a common pattern as cons that it is worth taking the first one. The first one was just drawing from the picture, and I really liked that I liked to draw a diagram, it really helps me understand a problem, even simple ones, when I don't have a diagram, they are confusing for me, but in addition to your diagram you also standardize the variables you label there and that lets someone else read your presentation and understand what the heck the L on the H even relates to, so write down your big equations first.
There's an equation for the constraint and an equation for

optimizing

the equation that you basically want to have that way, you might have several different constraints and a whole bunch of different variables, so at the end of the day when you're putting things in, you have that you get what you get when you put it in this optimization equation you get a function and it's in a variable and that's what you want, if you have a function of a variable then you can take its derivative, that's step number four, we take a derivative that gives us the critical numbers, and you remember that a critical number is a candidate for a maximum or minimum, it's not necessarily a maximum, it's not necessarily a minimum, but it is a candidate so you can try to classify all these critical numbers and the endpoints we want to find global max where is the global min Your model is interested in a relative max or minimum, which would depend on your application, but use the first derivative test either way.
The second derivative test checks the endpoints, whatever you need to do to be able to decide if it's a maximum or a minimum, and the last point is a bit beside the point, it seems really obvious that I'll call it answering the question, and the reason I'm asking this here is that different versions of the problem will often ask different kinds of things, in this case I asked for the dimensions, which meant I wanted both the L and the H. Sometimes the question might just ask about the L, or maybe you would ask what

volume

you're getting, not the dimensions of the So always read the issue carefully and answer exactly the question that posed the issue
Trending