1 00:00:00,012 --> 00:00:04,807 [music]. Let's suppose that you work for a soup 2 00:00:04,807 --> 00:00:12,999 company, and they've given you a task, a task to design a soup can that can hold 3 00:00:12,999 --> 00:00:20,104 300 times pi cubic centimeters of soup. This is really an optimization problem. 4 00:00:20,104 --> 00:00:25,090 Your goal is to design a soup can using as little metal as possible, which 5 00:00:25,090 --> 00:00:28,972 nevertheless holds 300 pi cubic centimeters of soup. 6 00:00:28,973 --> 00:00:33,789 Okay, less dramatically, you're really being asked to produce a cylinder with a 7 00:00:33,789 --> 00:00:36,837 given volume. I want the volume of the thing to be 300 8 00:00:36,837 --> 00:00:41,138 pi cubic centimeters, and I want you to minimize the surface area, right. 9 00:00:41,138 --> 00:00:43,769 This thing's got a side and a top and a bottom. 10 00:00:43,769 --> 00:00:47,379 And I want you to minimize the surface area for a given volume. 11 00:00:47,379 --> 00:00:52,633 Probably helps to review a little bit of the geometry that goes into this problem. 12 00:00:52,633 --> 00:00:57,603 Here's a picture of a cylinder right, and it's got some height that I'll be calling 13 00:00:57,603 --> 00:01:01,814 h and a radius that I'll call r. And the volume of this cylinder is pi r 14 00:01:01,814 --> 00:01:04,836 squared h. All right, pi r squared is the area, and 15 00:01:04,836 --> 00:01:09,670 I'm multiplying by h to compute the volume of this disk dragged through space. 16 00:01:09,670 --> 00:01:15,936 What's the surface area of this cylinder? Well 2 pi r squared counts the surface 17 00:01:15,936 --> 00:01:22,209 area of the top disk and the bottom disk. Each of those has pi r squared for their 18 00:01:22,209 --> 00:01:25,873 area. And the curved part of this cylinder has 19 00:01:25,873 --> 00:01:28,782 area 2 pi r h. Okay, let me draw a picture of my soup 20 00:01:28,782 --> 00:01:31,038 can. So, yeah, here's a little picture of our 21 00:01:31,038 --> 00:01:33,935 soup can, and you see I've labeled the relevant sizes. 22 00:01:33,935 --> 00:01:37,932 The height of the can and the radius of the can is what I'm trying to figure out. 23 00:01:37,932 --> 00:01:40,072 Now what is it that I'm trying to optimize? 24 00:01:40,072 --> 00:01:42,241 So I'm trying to minimize the surface area. 25 00:01:42,241 --> 00:01:46,267 Here's that formula for the surface area of this can, and I want that surface area 26 00:01:46,267 --> 00:01:49,361 to be as small as possible. Of course, if I wanted to make that 27 00:01:49,361 --> 00:01:53,236 surface area as small as possible, I just make a really tiny soup can, right? 28 00:01:53,236 --> 00:01:57,484 But there's a constraint. Well, the constraint is that soup can has 29 00:01:57,484 --> 00:02:02,425 to hold 300 pi cubic centimeters of soup. So I'm trying to make this quantity, the 30 00:02:02,425 --> 00:02:07,235 surface area, as small as possible, subject to the constraint that the volume 31 00:02:07,235 --> 00:02:11,498 of the soup can is in fact 300 pi. And I should also point out that the 32 00:02:11,498 --> 00:02:15,826 radius and height had better be positive. Otherwise, kind of nonsense. 33 00:02:15,826 --> 00:02:19,546 This thing that I'm trying to optimize really involves two variables. 34 00:02:19,546 --> 00:02:23,571 It involves the height and also a radius. Yeah this is some bad news, right? 35 00:02:23,571 --> 00:02:27,991 The quantity that I'm trying to minimize involves 2 variables, involves r and h. 36 00:02:27,991 --> 00:02:32,617 And calculus, as we've been setting it up, only works for understanding how a single 37 00:02:32,617 --> 00:02:38,015 variable changing affects something else. So I need to rewrite this function of two 38 00:02:38,015 --> 00:02:41,126 variables as a function of a single variable. 39 00:02:41,126 --> 00:02:46,375 So, here we go, let's take this constraint and let's solve for h, all right? 40 00:02:46,375 --> 00:02:51,810 And if I start doing that, well, h is 300 pi over pi r squared and I can cancel 41 00:02:51,810 --> 00:02:56,187 these, pis and I'm just left with h is 300 over r squared. 42 00:02:56,187 --> 00:03:02,657 So, for a given radius, this is how tall I need to make the soup can to guarantee 43 00:03:02,657 --> 00:03:07,421 that the soup can holds 300 pi cubic centimeters of soup. 44 00:03:07,421 --> 00:03:14,294 Now, once I've got an equation for h in terms of r, I can use this to rewrite the 45 00:03:14,294 --> 00:03:18,365 thing I'm trying minimize just in terms of r. 46 00:03:18,365 --> 00:03:23,489 So, here we go. The thing I'm trying to minimize now is 2 47 00:03:23,489 --> 00:03:30,769 pi r squared plus 2 pi r times h, but h, in order to satisfy the constraint, is 300 48 00:03:30,769 --> 00:03:35,342 over r squared. And now I can simplify this a little bit 49 00:03:35,342 --> 00:03:39,247 further, right? I've got an r and an r squared here. 50 00:03:39,247 --> 00:03:43,066 So I can get rid of this square and get rid of this r. 51 00:03:43,066 --> 00:03:48,896 And then, the thing I'm trying to minimize is 2 pi r squared plus 2 pi times 300, 52 00:03:48,896 --> 00:03:54,140 which is 600 pi divided by r. So this is the quantity that I'm trying to 53 00:03:54,140 --> 00:03:57,937 minimize. Now that I've got this thing written as a 54 00:03:57,937 --> 00:04:03,674 function of a single variable, I'm really tempted to do the fifth step here. 55 00:04:03,674 --> 00:04:07,289 Apply calculus, right? If I got a function of a single variable, 56 00:04:07,289 --> 00:04:10,164 I could differentiate. Find the critical points. 57 00:04:10,164 --> 00:04:12,745 Figure out where the maxima and minima occur. 58 00:04:12,745 --> 00:04:17,307 But before we dive in to the calculus side of this problem, let's just think about 59 00:04:17,307 --> 00:04:20,091 some creative solutions to the soup can issue. 60 00:04:20,091 --> 00:04:24,610 For example, here's a little model of a situation. 61 00:04:24,610 --> 00:04:31,112 Here's the sun, here's Earth and here is my new plan for soup cans, right? 62 00:04:31,112 --> 00:04:35,223 Really big radius, really tiny height, right? 63 00:04:35,223 --> 00:04:41,895 Even if r is you know, 100 million miles, that can make h small enough so that this 64 00:04:41,895 --> 00:04:47,287 soup can contains exactly 300 pi cubic centimeters of soup. 65 00:04:47,287 --> 00:04:50,493 [laugh]. Well, what's the surface area in this 66 00:04:50,493 --> 00:04:53,510 case? Is this a really good choice for a soup 67 00:04:53,510 --> 00:04:56,674 can, in, in terms of surface area? No, right? 68 00:04:56,674 --> 00:05:01,960 If I make r really big, admittedly, this quantity, 600 pi over r, which is 69 00:05:01,960 --> 00:05:06,491 measuring the curved area, that'll end up being very small. 70 00:05:06,492 --> 00:05:12,148 Right, but 2 pi r squared when r is really big, this quantity, the amount of metal in 71 00:05:12,148 --> 00:05:15,650 the top and the bottom of the can is enormous right? 72 00:05:15,650 --> 00:05:18,585 So this is not a great choice for soup cans. 73 00:05:18,585 --> 00:05:23,292 So having the soup can shaped like a giant pancake is a terrible idea. 74 00:05:23,292 --> 00:05:27,989 But what if I went the other way? Here is a piece of wire, what if I made 75 00:05:27,989 --> 00:05:31,990 the soup can really long and thin like this piece of wire? 76 00:05:31,991 --> 00:05:36,012 Well, here is that long, thin wire-shaped soup can, right. 77 00:05:36,012 --> 00:05:41,364 Even if I make the radius of my cylinder very small, if I make it long enough, then 78 00:05:41,364 --> 00:05:45,594 the soup can does hold 300 pi cubic centimeters of soup, right? 79 00:05:45,594 --> 00:05:50,550 And the question is, is this a really smart choice to minimize surface area, and 80 00:05:50,550 --> 00:05:55,251 no, it turns that if r is really small, that's really not all that helpful. 81 00:05:55,251 --> 00:05:58,991 Yeah, if r is really small, then 2 pi r squared is really small. 82 00:05:58,991 --> 00:06:02,828 I don't need very much metal on the top and the bottom of the soup can. 83 00:06:02,828 --> 00:06:07,175 But then 600 pi over r, which is how much metal is in the curved part of the soup 84 00:06:07,175 --> 00:06:09,578 can, this quantity is going to be enormous. 85 00:06:09,578 --> 00:06:13,234 So really, we see that these two things are competing against each other. 86 00:06:13,234 --> 00:06:17,209 A small value of r isn't really very helpful, and a really big value of r isn't 87 00:06:17,209 --> 00:06:20,272 very helpful. So our creative solutions didn't quite pan 88 00:06:20,272 --> 00:06:22,013 out. Let's do some calculus now. 89 00:06:22,013 --> 00:06:26,079 Okay, so I'm going to think of this surface area as a function of a single 90 00:06:26,079 --> 00:06:29,512 variable, r, right? Here it is, f of r is 2 pi r squared plus 91 00:06:29,512 --> 00:06:33,831 600 pi over r, and I'm going to differentiate, and here's the derivative. 92 00:06:33,831 --> 00:06:38,257 All right, I differentiate 2 pi r squared to get 4 pi r, and I differentiate this 93 00:06:38,257 --> 00:06:41,109 number divided by r. Well, 1 over r is negative 1 over r 94 00:06:41,109 --> 00:06:42,771 squared. That's its derivative. 95 00:06:42,771 --> 00:06:45,952 And the I multiply by 600 pi. So this, is the derivative of this. 96 00:06:45,952 --> 00:06:49,929 Now, I need to find the critical points. Now, remember what critical points are, 97 00:06:49,929 --> 00:06:52,322 right? Critical points are where the derivative 98 00:06:52,322 --> 00:06:55,145 vanishes, or where the derivative doesn't exist, right? 99 00:06:55,145 --> 00:06:58,896 The function's not differentible. This function, that was differentible on 100 00:06:58,896 --> 00:06:59,969 its domain. All right? 101 00:06:59,969 --> 00:07:03,618 You might worry when r is equal to 0. But that's not even in the domain of the 102 00:07:03,618 --> 00:07:07,203 original function. So I don't have to worry about that. 103 00:07:07,203 --> 00:07:12,793 All right, now we've gotta figure out, when is the derivative equal to zero, so 104 00:07:12,793 --> 00:07:17,606 I'm trying to solve this. I'll add 600 pi over r squared to both 105 00:07:17,606 --> 00:07:21,424 sides. So 4 pi r must be 600 pi over r squared in 106 00:07:21,424 --> 00:07:26,680 that case. I'll multiply both sides by r squared, and 107 00:07:26,680 --> 00:07:33,740 I'll divide both sides by 4 pi. And I'll get r cubed is 600 pi over 4 pi. 108 00:07:33,740 --> 00:07:39,552 The pis cancel, and 600 over 4 is 150, so r cubed must be 150. 109 00:07:39,552 --> 00:07:44,648 In other words, r is the cube root of 150. This is the only critical point. 110 00:07:44,648 --> 00:07:48,549 Now that I found the critical points, what about the end points? 111 00:07:48,549 --> 00:07:52,760 Well, the end points actually aren't really valid in this situation. 112 00:07:52,760 --> 00:07:55,857 One of the end points might be when r is equal to zero. 113 00:07:55,857 --> 00:07:58,569 But in that case, the soup can has no volume. 114 00:07:58,569 --> 00:08:03,102 So I can't satisfy the constraint. The same thing happens when h is equal to 115 00:08:03,102 --> 00:08:05,989 zero. In that case, the soup can, again, has no 116 00:08:05,989 --> 00:08:08,836 volume. So I can't satisfy the constraint. 117 00:08:08,836 --> 00:08:13,444 So, I don't have to worry about the end points, because they're not in the domain 118 00:08:13,444 --> 00:08:17,137 that I'm considering. But I do have to worry about the limiting 119 00:08:17,137 --> 00:08:20,453 behavior when r is really big or when r is really small. 120 00:08:20,453 --> 00:08:23,905 So, we actually already handled the limiting behavior. 121 00:08:23,906 --> 00:08:26,638 Right? We've already considered the situation 122 00:08:26,638 --> 00:08:30,281 when r is really small. That's the situation where the soup can's 123 00:08:30,281 --> 00:08:33,791 like a long, thin wire. And we checked, in that case the surface 124 00:08:33,791 --> 00:08:35,218 area is enormous. Right? 125 00:08:35,218 --> 00:08:39,644 If r is small enough, the surface area can be as large as you like, and still hold 126 00:08:39,644 --> 00:08:44,296 300 pi cubic centimeters of soup. We also consider the situation where r is 127 00:08:44,296 --> 00:08:48,152 really large, right? When r was really big ,we had this sort of 128 00:08:48,152 --> 00:08:52,758 flat pancake shape for the soup can. And in that case, we again saw that if r 129 00:08:52,758 --> 00:08:57,346 is big enough, the surface area can be made as large as you like, while still 130 00:08:57,346 --> 00:09:01,371 having that soup can contain 300 pi cubic centimeters of soup. 131 00:09:01,371 --> 00:09:04,983 So, the only thing that really remains is this critical point. 132 00:09:04,983 --> 00:09:09,477 And if you plug this critical point into the original function, you get this as the 133 00:09:09,477 --> 00:09:13,833 resulting surface area, if you build your soup can with a radius of cube root 150 134 00:09:13,833 --> 00:09:18,321 centimeters, and your height, whatever it has to be in order to guarantee that the 135 00:09:18,321 --> 00:09:21,451 soup can contains 300 pi cubic centimeters of soup. 136 00:09:21,452 --> 00:09:25,608 So how does this turn out? What's the best shape for a soup can? 137 00:09:25,608 --> 00:09:30,427 So the best case for your soup can is to use a radius of the cube root of 150. 138 00:09:30,427 --> 00:09:35,340 And you can see that from this graph. Here, I'm graphing the surface area for a 139 00:09:35,340 --> 00:09:39,958 given radius of soup can containing 300 pi cubic centimeters of soup. 140 00:09:39,958 --> 00:09:43,701 And you can see that the graph goes up towards infinity here. 141 00:09:43,701 --> 00:09:48,119 And up towards infinity there. And that's exactly the situation when the 142 00:09:48,119 --> 00:09:52,258 radius is very small or very big. The derivative is negative and then 143 00:09:52,258 --> 00:09:53,387 positive. Right? 144 00:09:53,387 --> 00:09:56,242 The function's decreasing and then increasing. 145 00:09:56,242 --> 00:10:01,724 So this value really is a local minimum. And in fact, this is the global minimum of 146 00:10:01,724 --> 00:10:03,444 this function. Right? 147 00:10:03,444 --> 00:10:07,633 And that's why this is the best choice for your soup can. 148 00:10:07,633 --> 00:10:11,638 Is this really a legitimate application of calculus? 149 00:10:11,638 --> 00:10:14,853 Well, here's a task to think about. Right? 150 00:10:14,853 --> 00:10:18,068 Go out, find some soup cans and check them. 151 00:10:18,068 --> 00:10:23,948 Are they really building soup cans in order to minimize the amount of metal that 152 00:10:23,948 --> 00:10:28,991 it takes to manufacture the can? Or perhaps there's other issues involved 153 00:10:28,991 --> 00:10:31,655 in manufacturing soup cans. .