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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 life application most of what you are trying to do in real life is optimize for different things you want to have the maximum amount of money your bank account you want to use the minimum cost of materials you want to minimize energy in your system you want to minimize the heat in your system whatever it is is a maximization problem or a minimization problem so this is one of the reasons everyone has to learn calculus and STEM disciplines because the need for

Optimizing

things is all over the place so in this particular section of our calculus course we're going to have a couple of different examples where we figure out a method of using the tools we've already developed to apply them to different optimization problems now in this first optimization problem i'm going to see i'm going to talk about a box is a box with a square bottom so the bottom is a perfect square then go up a little height and has no top but has a bottom now the first step of most optimization problems is to draw a picture, we want to put all the labels for our variables and just have a very clear picture of what is happening , so let's draw a box there we have my kind of terrible diagram ok if you need it to get a clearer picture here I'll let my helper show you what this box looks like ok James let's teach our students something calculus, so first of all, what are we going to say?
ex optimizing the volume of a box with fixed surface area
Let's call this the length and we'll call this down here the height I know let's rotate so in this case I have a square box and what I mean by this square base is that this bottom and this side are equal that is what it means to be a square so let's say maybe I'll call this bottom length L and there's actually a lot of ELL's there's an L there and there and they're anywhere along the bottom is going to be an L and they're also there they're up along the top there are pixies everywhere but we're all going to leave one and then I'm going to also do the height of this.
ex optimizing the volume of a box with fixed surface area
I'll call it 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 some equations that reflect this particular. box now the first equation that I'm going to write that relates to this box is about the constraint the constraint in this problem is that I only have 12,000 square centimeters of material to work with out of those 12,000 square centimeters of material that I have to make the base I have to make the four sides different, so the total

area

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

area

. my constraint my 12,000 and then this equals L squared What is l squared? It represents that it is the base of my particular box. You have an L here and then an L there, so the

area

of ​​that square at the bottom is l squared. I also have this thing called o LH Y for lunch well let me take let's say this panel right here this panel has a length of L at the bottom and then a height of H so this rectangle over there will have an

area

of ​​L times H, but I have four of these sides. they're all identical so that's four times L times H that's my restrictive equation the next equation I want to write is what I call my optimization equation because what I want to optimize what I want to be as large as possible like my

volume

is the blue one so they have a different formula for my optimization equation and here it is it's the

volume

of l squared H this is the

volume

of a box where the base has width and length equal to same thing go to L and then height H then L squared H now we know what I want to go and optimize a particular function of one variable takes the derivative and sets it equal to zero but the problem is that I can't directly take the derivative of this

volume

and set it equal to zero because with respect to which there are multiple different variables there is an L here and an H here if fol were a function of a single variable then you could use our methodology that we have seen before take its derivative ig set it to zero so i want to take this

volume

that has two variables and i want to write it in terms of a single variable how can i do that? my trick is this i'm going to take the restrictive equation i'm going to try to plug it into my optimization equation i'm trying to get rid of one of the variables i have to try to get rid of in particular if i look at this i can get rid of L or H but just by its side it's only there once and the L appears in two different places so it makes sense to me let's make H equal to blah and we'll take the H and put it there then I'll just be left with LS.
ex optimizing the volume of a box with fixed surface area
I'll show you how to do this, so first of all, let's take a constraint equation, let's rewrite it. I have solved it. this for H. I brought the 1200 rearranged. There's my formula for H. Now I'll take that H. I'll put t. hat down in the

volume

and where there used to be an age now there's this big messy thing okay so I've got this formula let's get this out okay I got that looks pretty good but I'm going to clean it up a bit like this which I'm going to go and just take the l squared here and the L cancels that out I just get an L and then the l squared over the L is 1 L times an l squared which gives me my nice L it's a little bit cleaner , well progressing because now I can tell I can tell because it's just a function of a variable l so that's pretty cool so let's take the derivative of this it's just a polynomial and now we can do that so its derivative is the derivative of the

volume

with respect to the length it's fine 300 the L reduces to a cube of 1 L reduces to 3 L squared and what I'm doing is setting it equal to zero.
ex optimizing the volume of a box with fixed surface area
I want to find the critical number. I want to find out where there is a candidate. to be a maximum or minimum Asst ok relatively simple equation this is just a quadratic i can take the 3l squared move it to the other side take the three thousand multiply it by 4/3 square root and what do i get i get that L is 20 root 10 centimeters by the way because the square is technically plus or minus 20 root 10 centimeters but let's think about this physically the length of a negative number to build a block doesn't make any sense so I'll just take the positive root so this is my critical number and this is the value of L where I am going to have a maximum or at least I am completely sure that it is a critical number, its derivative is zero, but it gives a maximum, it could give a minimum, there are a couple of things that I want check the first point to make is that the endpoints under consideration none of them physically make a box with

volume

as i certainly can't have negative values ​​of L that doesn't make any sense a lon Negative gitude doesn't make sense but if L is zero then this is a box with no base at all which has no

volume

if L is infinities this is like an infinite plane at the bottom which would force my height to be zero so again would have no

volume

so the endpoints don't make any sense now i can reason that because the maximum has to occur at an endpoint or at a critical number and clearly they are points where this has positive

volume

we can make a chart with the positive number the maximum has to occur at this critical number since it's not occurring at the other endpoints The way I can reason through this is the first test primitive so if I actually look at what this derivative is if I do that my L to be a larger number because of this minus sign here my derivative becomes negative when L is larger if my L is smaller than this particular critical number then I am subtracting less and so I have a positive number, so what happens to my function goes from increasing to decreasing.
ex optimizing the volume of a box with fixed surface area
I have a maximum through the first derivative test, so this is all to say that the critical number the computer really is the maximum, now we have this value of l, this L is 20 root of 10 centimeters, but what? what did i ask in the problem? the age so let me mention just that L here and then I'll add a note that we had a formula for what the H was the formula for H was that it was just this thing we had previously told me H in terms of L so if I take that particular value of L and load it for my age, now I have a new value of H, it's 10 roots, 10 centimeters, full stop, let's think about what this looks like physically.
I tried to draw it wrong, but at least relatively accurately to what the maximum

volume

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

volume

of this open box has a height that is half this length, okay? so now let's summarize the key steps we've done in each optimization In truth, one problem is a bit different from the others, but there is a common pattern that the steps against, it's worth taking the first one, it was just the drawing of the image and I liked it a lot.
I liked drawing a diagram. It really helps me. understanding a problem, even the simplest ones when I don't have a diagram, they confuse me, but in addition to your diagram, you're also standardizing the variables, you put the labels in there and that allows someone else to read your presentation and understand what the Heck, is the L in the H even referring so do that first second of all you want to write your big equations. There is an equation for the constraint and there is an equation for

optimizing

the equation. What you want to have in principle for the You might have multiple different constraints and a bunch of different variables so that at the end of the day when you plug things in, what you get when you plug it into that optimization equation is that y 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 so that's step number four we take a derivative which gives us the critical numbers and remember that a the critical number is a candidate to be a maximum or a minimum, it's not necessarily a maximum, it's not necessarily a minimum, but it's a candidate, so you can go and try to classify all those critical numbers and endpoints that we want to try to find where the global maximum where is the global minimum?
Your model is interested in a relative maximum or a relative minimum, it would depend on your application, but either way, use the first derivative test, the second derivative test, check the endpoints, whatever you need to do to being able to decide if it's a maximum or a minimum and the endpoint is a little bit less seems very obvious I'll call it answering the question and the reason I'm putting this here is that different versions of the p Roblem 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 may be just the L or maybe you ask what

volume

you get, not the dimensions of the. so always be sure to read the problem carefully and answer the exact question that caused the problem
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