Learn about Optimization with Calculus 2 Video

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Learn about Optimization with Calculus 2

This next problem is a bit of classic and I’ve actually got multiple requests to do some variation of this problem. So, it says, find the volume of the largest open box, open box that can be made from a piece of cardboard, 24 inches square. Somebody draw on my piece of cardboard. That’s not how I want it, I want to draw it filled in. So, my piece of cardboard is going to be a square. I draw it pretty big because I want you to visualize it. And it’s 24 -- what did they say, 24 inches square. So, it is 24 by 24 square, by cutting equals squares from the corners and turning up the side. So, what did they say? Well, they said we’re going to cut equal squares from the corner. So let’s say we cut a -- let me see if I can draw this well. So, I’m going to cut a square here and I’m going to cut equal size, square there. I want to draw this neatly. I’m going to cut it equal size, square there and then I cut it equal size, square there, and right there. So, I’m going to cut out those corners and then I’m going to fold up the sides. So you can visualize it. So those dotted lines are where I would actually fold the box, and then I would end up with an open box and let’s see what the dimensions would be. And that actually would the whole cracks to the problem is. So, how big of a square should I cut out in order to maximize the volume of the box? Well, these are equal sided squares that I’m cutting out. So, let’s say that each side is X. So if this is X, then this is X, this is X, this is X, this is X, this is X, this is X, this is X. And then what is going to be the volume of our box when we fold up everything. Let me see. So I’ll draw the base. So this dotted square right here, that’s going to be the base, right? I’ll draw it an angle. That’s the base. And where are the dimensions of the base? Well, this base is going to be the length of this side of the cardboard minus this X, minus this X, right? So, this base, right here, that is going to be 24 minus two X, right? It’s just the whole length minus this, minus that. That’s how long that is. And, of course this is going to be the same length, this is going to be that same length. All of these sides of the base are 24 minus two C. So, it’s 24 minus two X and then this side is also 24 minus two X. And then what is going to be the height of this open box that we are creating? So, let me draw the sides of it. That’s open and that means it doesn’t have a top to it. So we have an opening here, so you can kind of see the inside of the box. And what is this height? Well, when you fold this flops up, when you fold this flops up, the X, the height is X, right? If you can think about it, this side here -- this is really just visualization. This side right here, that you can view as thing right here. I can even do it in different color for you. You can view this side, this front facing side, is this right here. I’ll just do a couple, I don’t have to do all of the sides that I think, you could view this side as this side right here, right, it’s folded up and then that this backside is this backside and that side there and this is the base. Anyway, so that is the volume of our open box and we want to maximize it. So, how do we do that? Well, let’s write the volume as a function of X and then take us derivative, figure out where the derivative is zero and hope that that’s the maximum point and we’ll prove it by taking the second derivative and saying, now, we actually want it to be concave downwards, because concave downwards looks like that s, so that means you found the maximum point. So, we know that the volume has a function of X, so like drawing the Vs. The volume has a function of X is that side times -- you know, you could view it as depth times length times height. So, its X the height times the depth, 24 minus two X times the width. 24 minus two X, and let’s see if I can multiply this out. It’s probably the hardest part about this problem. So, X times -- what’s 24 times 24 it’s, I want to say it’s -- what is it four. I will show you. It’s kind of sad but I’ve made so many careless mistakes recently that I don’t want to make anymore and I want to get the right number. So, four times four is 16, four times two is eight, 96, zero, two times 24 is 48. six, nine plus 87, 10, 576. That’s what I thought it was. I should’ve memorize my times table up to 25. But anyway, 24 times 24 is 576. 576 and then we’ll have two time 24, so that’s 48 X but we do have them twice, so it’s minus 96 X, right, because two X times 24 -- minus two X times 24 then minus two X times 24, so it’s 48 plus 48, so minus 96 X and then finally, the last two plus four X^2, 4 X^2 and now we can multiply the X is out and we get the volume of the open box is – let me put the X term first. 4 X to the third, I’m just multiplying this out, minus 96 X^2 plus 576 X. Let me erase this. I should’ve been able do that in my head, but anyway. I’m not too proud to show that I didn’t know that. Let’s see, so let me erase that, let me erase that. And so, what do we want to do? We want take the derivative of this, figure out what X values. Do we have a zero slope and then test to see if those are maximum or minimum points. If we have a maximum point at that X value and if it’s a global maximum, then we have found the X value that optimizes the volume. And you might want to graph this and experiment with them and get more intuitive sense of it, but that’s really what we did when we found the minimum and maximum, we did concave upwards and downwards and all of that. So anyway, what’s derivative? V prime of X is equal to 12 X^2 -- this part is the fun part, 12 X^2 minus, was it 180, 192, 192 X plus 576. I know it’s hard to read. It’s hard to read for me. So let figure where this equal zero. So we want to know where 12 X^2, 12 X^2 minus 192 X plus 576 is equal to zero. And the easiest thing to do is here is just to divide the whole thing by 12, both side because we divide zero by 12. You still get zero, so you get X^2minus -- what’s 192 divided by 12? Well, 192 is 24 times 4, right, that’s 24 times four -- is that right? 24. No, no, 24 times 16 is 192. I have been making so many careless mistakes. 12 was 192, 1272, 16. I should’ve been able to -- anyway, minus 16 X plus -- well, this was 24 times 24.So, this is also, you could do -- this should be 12 times 48. That should be plus 48 is equal to zero. And we could -- 24 times 24 is 576. So 12 times 48, right. So now we just to have factor this. So, what two numbers that when I add then, I’d get minus 16 and then when I multiply them, I get 48, +48. Let’s see. Six times -- let me make sure I did t his. So, 12, 12 goes into 72, six times -- I just want to make sure I have the numbers right. As you can tell, I don’t do these problems ahead of time because I want you to see that I go through the pain as you do. So, let me make sure.12 goes into 576, 12 goes into 576, four times 48, 48 and you get 96. 12 goes into 9 – right, 48, 12. I shouldn’t have had to do that. After making so many careless mistakes and I haven’t eaten dinner yet. So anyway, maybe I’m just missing the easy factor, right, because I have -- what numbers? You get 12 times -- oh I actually I lust figure it out, 12 times 4? Sometimes when your brain is in calculus mode, the algebra too gets difficult. So, this is the same thing as X minus four times X minus 12 is equal to zero. So X is equal to four, whereat some type of minimum or maximum, where at some critical point and that X is equal to 12. Now, let me tell you something. You don’t even have to look at the secondary ahead of you. What happens when X is equal to 12? What will happen to the volume of the box? If X is equal to 12, what is the length of this base? It’s zero, right? Because the length of this base is 24 minus 2 X. So, this would be zero, this would be zero. So this is a point where we have a volume of zero. So you know that this is going to be a minimum point. And if you want to verify it, take the second derivative of the volume and evaluate the second derivative of the volume at 12 and you will realize that we are concave upwards and that X equals 12 is a minimum point. So, you already know in all probability, the answers X equals four. And if you really wanted to verify, you could take the second derivative at X four and make sure that we are concave downward at that point, right, because if we want a maximum point, we want to be at some place where the graph look like some thing like that. So we’re at our maximum point. So, what’s the second derivative? V prime, prime of X is equal to 24 X minus 192. And what’s 24 times four? It’s 96, right? So, C prime, prime with four is equal to 96, right, four times 24 minus 192 is equals minus 96. So the second derivative at this point is negative which means that we are concave downwards, which means that this is maximum point. And once again, I challenge you to find another value of X where you get a larger value. And just to get an intuition, how large will this box be? Well, if X is four then each side is 24 minus eight. This would be 16 by 16 by four. So that would be optimal volume for that box. Hope you found that useful and I will see you with the next video where I will do another optimization problem, and this one is especially fun. See you soon.

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This next problem is a bit of classic and I’ve actually got multiple requests to do some variation of this problem. So, it says, find the volume of the largest open box, open box that can be made from a piece of cardboard, 24 inches square.

Somebody draw on my piece of cardboard. That’s not how I want it, I want to draw it filled in. So, my piece of cardboard is going to be a square. I draw it pretty big because I want you to visualize it. And it’s 24 -- what did they say, 24 inches square. So, it is 24 by 24 square, by cutting equals squares from the corners and turning up the side.

So, what did they say? Well, they said we’re going to cut equal squares from the corner. So let’s say we cut a -- let me see if I can draw this well. So, I’m going to cut a square here and I’m going to cut equal size, square there. I want to draw this neatly. I’m going to cut it equal size, square there and then I cut it equal size, square there, and right there.

So, I’m going to cut out those corners and then I’m going to fold up the sides. So you can visualize it. So those dotted lines are where I would actually fold the box, and then I would end up with an open box and let’s see what the dimensions would be.

And that actually would the whole cracks to the problem is. So, how big of a square should I cut out in order to maximize the volume of the box? Well, these are equal sided squares that I’m cutting out. So, let’s say that each side is X. So if this is X, then this is X, this is X, this is X, this is X, this is X, this is X, this is X. And then what is going to be the volume of our box when we fold up everything. Let me see. So I’ll draw the base. So this dotted square right here, that’s going to be the base, right? I’ll draw it an angle. That’s the base. And where are the dimensions of the base?

Well, this base is going to be the length of this side of the cardboard minus this X, minus this X, right? So, this base, right here, that is going to be 24 minus two X, right? It’s just the whole length minus this, minus that. That’s how long that is. And, of course this is going to be the same length, this is going to be that same length. All of these sides of the base are 24 minus two C. So, it’s 24 minus two X and then this side is also 24 minus two X. And then what is going to be the height of this open box that we are creating?

So, let me draw the sides of it. That’s open and that means it doesn’t have a top to it. So we have an opening here, so you can kind of see the inside of the box. And what is this height? Well, when you fold this flops up, when you fold this flops up, the X, the height is X, right? If you can think about it, this side here -- this is really just visualization. This side right here, that you can view as thing right here. I can even do it in different color for you.

You can view this side, this front facing side, is this right here. I’ll just do a couple, I don’t have to do all of the sides that I think, you could view this side as this side right here, right, it’s folded up and then that this backside is this backside and that side there and this is the base.

Anyway, so that is the volume of our open box and we want to maximize it. So, how do we do that? Well, let’s write the volume as a function of X and then take us derivative, figure out where the derivative is zero and hope that that’s the maximum point and we’ll prove it by taking the second derivative and saying, now, we actually want it to be concave downwards, because concave downwards looks like that s, so that means you found the maximum point.

So, we know that the volume has a function of X, so like drawing the Vs. The volume has a function of X is that side times -- you know, you could view it as depth times length times height. So, its X the height times the depth, 24 minus two X times the width. 24 minus two X, and let’s see if I can multiply this out. It’s probably the hardest part about this problem. So, X times -- what’s 24 times 24 it’s, I want to say it’s -- what is it four. I will show you. It’s kind of sad but I’ve made so many careless mistakes recently that I don’t want to make anymore and I want to get the right number.

So, four times four is 16, four times two is eight, 96, zero, two times 24 is 48. six, nine plus 87, 10, 576. That’s what I thought it was. I should’ve memorize my times table up to 25. But anyway, 24 times 24 is 576. 576 and then we’ll have two time 24, so that’s 48 X but we do have them twice, so it’s minus 96 X, right, because two X times 24 -- minus two X times 24 then minus two X times 24, so it’s 48 plus 48, so minus 96 X and then finally, the last two plus four X^2, 4 X^2 and now we can multiply the X is out and we get the volume of the open box is – let me put the X term first. 4 X to the third, I’m just multiplying this out, minus 96 X^2 plus 576 X. Let me erase this.

I should’ve been able do that in my head, but anyway. I’m not too proud to show that I didn’t know that. Let’s see, so let me erase that, let me erase that. And so, what do we want to do? We want take the derivative of this, figure out what X values. Do we have a zero slope and then test to see if those are maximum or minimum points. If we have a maximum point at that X value and if it’s a global maximum, then we have found the X value that optimizes the volume.

And you might want to graph this and experiment with them and get more intuitive sense of it, but that’s really what we did when we found the minimum and maximum, we did concave upwards and downwards and all of that. So anyway, what’s derivative? V prime of X is equal to 12 X^2 -- this part is the fun part, 12 X^2 minus, was it 180, 192, 192 X plus 576. I know it’s hard to read. It’s hard to read for me.

So let figure where this equal zero. So we want to know where 12 X^2, 12 X^2 minus 192 X plus 576 is equal to zero. And the easiest thing to do is here is just to divide the whole thing by 12, both side because we divide zero by 12. You still get zero, so you get X^2minus -- what’s 192 divided by 12? Well, 192 is 24 times 4, right, that’s 24 times four -- is that right? 24. No, no, 24 times 16 is 192. I have been making so many careless mistakes. 12 was 192, 1272, 16. I should’ve been able to -- anyway, minus 16 X plus -- well, this was 24 times 24.So, this is also, you could do -- this should be 12 times 48. That should be plus 48 is equal to zero. And we could -- 24 times 24 is 576. So 12 times 48, right.

So now we just to have factor this. So, what two numbers that when I add then, I’d get minus 16 and then when I multiply them, I get 48, +48. Let’s see. Six times -- let me make sure I did t his. So, 12, 12 goes into 72, six times -- I just want to make sure I have the numbers right. As you can tell, I don’t do these problems ahead of time because I want you to see that I go through the pain as you do. So, let me make sure.12 goes into 576, 12 goes into 576, four times 48, 48 and you get 96. 12 goes into 9 – right, 48, 12. I shouldn’t have had to do that. After making so many careless mistakes and I haven’t eaten dinner yet.

So anyway, maybe I’m just missing the easy factor, right, because I have -- what numbers? You get 12 times -- oh I actually I lust figure it out, 12 times 4? Sometimes when your brain is in calculus mode, the algebra too gets difficult. So, this is the same thing as X minus four times X minus 12 is equal to zero. So X is equal to four, whereat some type of minimum or maximum, where at some critical point and that X is equal to 12.

Now, let me tell you something. You don’t even have to look at the secondary ahead of you. What happens when X is equal to 12? What will happen to the volume of the box? If X is equal to 12, what is the length of this base? It’s zero, right? Because the length of this base is 24 minus 2 X. So, this would be zero, this would be zero. So this is a point where we have a volume of zero. So you know that this is going to be a minimum point.

And if you want to verify it, take the second derivative of the volume and evaluate the second derivative of the volume at 12 and you will realize that we are concave upwards and that X equals 12 is a minimum point.

So, you already know in all probability, the answers X equals four. And if you really wanted to verify, you could take the second derivative at X four and make sure that we are concave downward at that point, right, because if we want a maximum point, we want to be at some place where the graph look like some thing like that. So we’re at our maximum point.

So, what’s the second derivative? V prime, prime of X is equal to 24 X minus 192. And what’s 24 times four? It’s 96, right? So, C prime, prime with four is equal to 96, right, four times 24 minus 192 is equals minus 96. So the second derivative at this point is negative which means that we are concave downwards, which means that this is maximum point.

And once again, I challenge you to find another value of X where you get a larger value. And just to get an intuition, how large will this box be? Well, if X is four then each side is 24 minus eight. This would be 16 by 16 by four. So that would be optimal volume for that box.

Hope you found that useful and I will see you with the next video where I will do another optimization problem, and this one is especially fun. See you soon.

Transcription by: Scribe4you Transcription Services