Stationary Points: Exercise 4

by Batool Akmal

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    00:01 The final question on this exercise is asking you for a practical use of stationary points and the things that we’ve just been looking at. Now remember that stationary points in itself as a topic is extremely practical. It’s used extensively in things like the study of hydrology or engineering, and for the medics and chemists amongst you, for things like molecular modeling or you're doing things like confirmation analysis where you look at bonds changing or rotations or the positions changing of molecules.

    00:34 Let's have a look at this question and explain a little bit about what actually is happening.

    00:41 We have the surface area of a closed rectangular box. If I try to model that here, we have a closed rectangular box here, so something like that. It’s telling us that the surface area of this, so the surface area basically means the areas of all surfaces is 108 m². It tells you that the length of the box is 2x meters.

    01:10 We have the length of the box, let’s just call that 2x there. Then the width is x m and the height is y m.

    01:18 So, if we say that this width here is x meters and the height is y meters, so we have all the dimensions of this box. It says calculate the maximum volume and the associate x value. What we’re actually doing is looking for how to maximize the volume of this box. What dimensions should we use to maximize or minimize the volume? As you can see already, this is a very useful question because it’s telling you dimensions that would maximize or minimize volume or the area of a surface or of an object.

    01:55 So, let’s start by looking at the surface area. They’ve told us that the surface area is 108 m².

    02:04 Now, remember that the surface area means the area of all surfaces added together. So, if we start with one surface at a time, so let’s just look at this surface here. Let’s just call this the front face here.

    02:17 We’ve got 2x and then the height of this, we’ve already been told is y. We have the first surface.

    02:23 We can say that that is 2x multiplied y. But remember there’s two of them. There’s this surface here.

    02:30 Then we’ve also got the same thing back here. So there's two of the same surfaces.

    02:36 So because there’s two of them, we can just multiply this with 2. Let’s look at our second surface.

    02:42 We can say if I use a different color, we can say that the second surface is this surface here.

    02:49 Remember that there’s two of them, so we’ll have another of those surfaces here. See if we can work out what the area of that surface would be, remember that’s just going to be x multiplied by y.

    03:03 So we’re just looking at the areas. We can say that’s x multiplied by y but again, there are two of them.

    03:09 So remember that there are two blue surfaces. Lastly, we have two surfaces left. We’ve got the base at the bottom. We've got this over here if you try and look inside the box and then at the top of the box there.

    03:24 We’re dealing with all three surfaces but essentially, there are six surfaces here. So, if we talk about that area now, so where if we just look at the top here and imagine that’s the lid. So this part here, this has a length of 2x from here and then it has a width of x. So, this will be 2x multiplied by x.

    03:49 Remember again that there are two of them. So, this is our surface area as an algebraic expression.

    03:57 Let’s just tidy this up. So that gives me 4xy + 2xy + 4x². Remember that they’ve already told us in the question that this is 108 m². We can just write it as a 108. There are two terms we can add here.

    04:15 So, we can now say that this is 6xy + 4x² = 108. We can leave that for now because we don’t really know what we need out of this equation. We can just call this equation 1. The question, remember, is asking you to calculate the volume. Let’s see if we can come up with an expression for the volume as well.

    04:37 So the volume, simply put, is just length times width times height. So it’s length times width times height.

    04:46 Let’s get the dimensions for our box. The length is 2x, the width is x, and the height is y.

    04:54 So we have 2x multiplied by x times y. Before I simplify this equation a little bit, if I just see that I have an xy here and an xy here. So if I just think one step further, what I could do is rearrange equation 1 to give me xy and then I could substitute that here to calculate an expression for the volume. If I aim to do that in this case, if I do 6xy = 108 – 4x², once I have done this, I can then get 6xy by itself. I can do, that’s 108. That’s 180. So if I just put 108 here, so (108 – 4x²)/6.

    05:42 I've divided it with this 6 here but both terms are being divided by the 6. I can simplify this a little bit to say xy is (108/6) - (4/6)x² just to see if they divide and if I can make this a little bit simpler.

    05:58 Got xy, that here will give me 6 divided 10 goes into it once and then I’ve got 4 here, so that gives me 18.

    06:08 Then I’ve got (2/3)x². So I now have xy by itself which is what I was aiming to do. So I have this xy here If I now call this equation, let’s call this equation 2 or 1 altered. But if I now sub 2 into volume, so I’m substituting this into the volume equation, I can say that the volume is 2x.

    06:37 Remember, I'm replacing this xy with this xy here. So that xy is going to be 18 – (2/3)x².

    06:49 So the volume, when I multiply this through is 36x – (4/3)x³. So all I’ve done is I’ve multiplied this through.

    07:01 So 2 times 18 gives me 36. The x stays there. Then I multiply just the 2 with the 2 here at the top.

    07:09 So that gives me (4/3)x³. Now I have an expression for the volume. Coming back to the question, it says calculate the maximum volume. Now, maximum and minimum terms to mathematicians like us now means some form of differentiation. So remember where maximum comes from.

    07:31 Think of this previous lecture and think of where we’ve used the term maximum. We used it when we were calculating stationary points. Then to calculate the nature of the stationary points, we often said that if the second differential is less than zero, you will end up with a maximum point.

    07:48 It’s that kind of a question. We’re looking for the maximum point and we’re using this volume here.

    07:54 Let’s just do that on the side here. We’ve said that the volume is 36x – (4/3)x³. Remind yourself now.

    08:09 How do we calculate the maximum point? To calculate the maximum point, you firstly need the stationary point. We need to know what the stationary point is. We can then do the second differential.

    08:20 We can substitute the stationary point in to figure out whether it’s a maximum or minimum point.

    08:26 So for the first time, we’re now looking at volume equals to something. We’ve been dealing with y functions and all other types of different equations. But now, we’re dealing with a volume. Now if you think about this, we need to differentiate our volume. So we can now differentiate our volume and say dV/dx because we’re differentiating it with respect to x because of the x’s in here. When we start to differentiate, 36x just goes to 36. This, bring the power down, decrease the power by one.

    09:00 I can bring my power 3 to the front. I have (4/3)x². This cancels out leaving me with dV/dx = 36 – 4x².

    09:16 Think about this now. How do you calculate the stationary points? Hopefully it’s all coming back to you.

    09:23 Stationary points occur when dy/dx = 0. In our case, when dV/dx = 0. So we’d want to know that dV/dx = 0 at stationary points. All we do is we equal this equation to zero. We can now solve this.

    09:43 You can say 36 = 4x² when I take it over. 36/4 = x² which gives me 9 = x². Square root your 9 to give you x, to give you ±3. So we’ve got two x values, this x value here. We’re saying that either this x value is 3 or the x value is -3. Now this is fairly easy to conclude to. You know that a length can’t be negative. So if we have a choice to choose between a length, +3 or -3, we already know that, there should have been an equal in there, but that x cannot be -3. So it has to be positive 3.

    10:30 We’re sort of answering the question anyway. We’re saying that this has to be the only value of x.

    10:36 Therefore, it must be a maximum. But let’s just prove it mathematically and make sure that we’re fully convinced that x = 3 is going to give this shape, this box a maximum value. In order to do that, we can now do our second differential, d²V/dx². We’re using this gradient here.

    10:58 When we differentiate this, the 36 disappears because it’s a constant. I now end up with -8x.

    11:05 Substitute x = 3 in there. You can do -8(3) to give you an answer of -24. This is a negative answer.

    11:16 So -24 is less than zero. Remember what less than zero means. It looks like this, so this is a maximum point.

    11:24 If you were designing something or if you were looking at things like molecular movement or so on and you’re looking for the bond length and so, and you’re looking for the maximum or the minimum possible length, you would use this. And you would use this fact at the very end once you know the expressions to actually be able to find whether this would give you a maximum or a minimum.

    11:48 It really is quite amazing to think that math can do this for you. It can tell you that this dimension of the box will give you the maximum value and anything else won’t be the maximum. So, in order to get the maximum volume in this particular box which has a surface area of 108, you would need to use an x value of 3.

    12:10 This will obviously become 2 times 3, so that gives you a length of 6. Then you can calculate your y by putting it back into this equation here. So the answer to the very last bit of this question, it says calculate the maximum volume and the associated x value. We found the x value, we’ve said that that x value is 3. But we can also now do the very last bit which actually asks us to calculate the maximum volume of this shape when you have a dimension x = 3. So for that, let’s go back to our volume expression here which we derived earlier. We said that volume is 36x – 4x³.

    12:51 We can find that out when x = 3. If we just put that in, we can say that the volume is 36(3) – (4/3) x or (3)³ in our case. If we make this a little bit easier, we’ve got 36 times 3 which gives you 108.

    13:16 This 3, one of these 3’s will cancel with one of them. So we’ve just got 3² here. So we’ve got -4(3)². 3² is 9.

    13:27 So that gives me -36 giving you an answer of 72 m³ because we’re dealing with volume.

    13:35 We have now found the maximum volume of this cube, also the associated maximum x value which is 3.

    About the Lecture

    The lecture Stationary Points: Exercise 4 by Batool Akmal is from the course Stationary Points.

    Included Quiz Questions

    1. S = 2(LW + WH + HL) , V = LWH
    2. S = 2(L² + W² + H²) , V = LWH
    3. S = 4(LW + WH + LH) , V = L³ + W³ + H³
    4. S = 2(LW + WH + LH) , V = 2LWH
    5. S = 2(L² + W² + H²) , V = L³ + W³ + H³
    1. S = 2x² + 4xh , V = x²h
    2. S = x² + 2xh , V = xh²
    3. S = 4x² + 4xh , V = x³
    4. S = 8xh , V = x²h
    5. S = 4x² + 4h² , V = hx² + xh²
    1. V = (75/2)x − x³/2
    2. V = (75/2)x − x³
    3. V = 75x − 2x³
    4. V = x³
    5. V = (75/2)x² − x³/2
    1. x = 5 m
    2. x = 10 m
    3. x = 6 m
    4. x = 5/2 m
    5. x = 7 m
    1. V = (1/3)(48x - 4x³)
    2. V = 16x - 4x³
    3. V = (1/3)(48x - 2x³)
    4. V = x³
    5. V = 16 - (4/3)x³
    1. x = 2 , Max volume = 64/3
    2. x = 1 , Max volume = 44/3
    3. x = 3 , Max volume = 12
    4. x = 3/2 , Max volume = 6
    5. x = 1/2 , Max volume = 47/6

    Author of lecture Stationary Points: Exercise 4

     Batool Akmal

    Batool Akmal

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