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Optimization: box volume (Part 2)

Finishing up the last video by working through the formulas. Created by Sal Khan.

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  • blobby green style avatar for user steven4276
    But what are the odds that "12x^2 - 200x + 600" just happens to be in the correct format to use it in the quadratic formula? What if it were: "-200x + 600 = 0" or "12x^3 - 200x + 600 = 0"?
    (30 votes)
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    • old spice man green style avatar for user Leonardo Martinez
      Hi Steven. That is an interesting question. Actually it happens to be a quadratic by chance, or more precisely, because the people who invented that exercise wanted the solver to combine his knowledge in calculus and the quadratic formula.

      In many other cases it is not so simple to get an exact answer, and you would have to use the "graphing method" Sal explained or make some numerical approximations (either to maximize the function or to set the derivative equal to zero).
      (62 votes)
  • blobby green style avatar for user Jessie Bear
    Why did you use 20-2x to find the domain and not 30-2x?
    (23 votes)
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  • leaf grey style avatar for user Matthew Rogers
    I have an issue with a simular problem. My sheet of metal is given as 40cm x 80cm, and when I follow this method, I end up trying to square root a negative number. Which of course isn't possible, at least without moving into the realm of complex numbers which won't help with my problem.
    (6 votes)
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    • blobby blue style avatar for user First L
      Hi Matthew! I solved this problem for you (mostly for personal practice). I'm not sure how much it will help considering it's 8 years later.

      The following is in Markdown and LaTex. I recommend copying the following content and pasting it into https://stackedit.io/app for easier reading

      ## Problem
      We have a sheet of metal that is $40 \text{ cm} \times 80 \text{ cm}$ and we want to fold it such that we get the most volume out of it.

      Given that, this is what our equation looks like:

      V(x) &= x(40-2x)(80-2x)\\
      &= x(3200-80x-160x-4x^{2})\\
      &= x(4x^{2}-240x+3200)\\
      &= 4x^{3}-240x^{2}+3200x

      Derive the above to get:
      V'(x) &= 12x^{2}-480x+3200\\
      &= 4(3x^{2}-120x+800)

      We need to use the quadratic formula to figure out when $V'(x)=0$. Doing so will help us figure out where the absolute maximum and minimum are. Luckily, the highest degree in the polynomial is $x^3$ (which means there is only one maximum and minimum).

      x &= \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\\
      &= \frac{120 \pm \sqrt{(-120)^{2}-4 \cdot 3 \cdot 800}}{2(3)}\\
      &= \frac{120 \pm \sqrt{4800}}{6}\\
      &\approx 8.453, \cancel{31.547}\\
      & \boxed{\approx 8.453}

      So, you would need to fold $8.453 \text{ cm}$ inwards.

      Note that the $x$ values we found do *not* need to be changed any further because they both make the quadratic equal $0$.

      You might also wonder how I already knew $31.547$ wasn't the answer. The answer is simple. If you plug in $31.547$ into $V(x)$, you get:
      $$V(31.547) = 31.547 \cdot \Big(40-2(31.547)\Big) \cdot \Big(80-2(31.547)\Big)$$

      Without solving any further, it's obvious that $2(31.547) \approx 63$, which would make it one parentheses negative. Meanwhile, all other terms are positive, so it would make $V(x) <0$ too.

      So, all we need to do is find $V(8.453)$:
      V(x) &= x(40-2x)(80-2x)\\
      V(8.453) &= 8.453(40-2\cdot 8.453)(80 - 2 \cdot 8.453)\\
      &= 8.453(40-16.906)(80 - 16.906)\\
      &= 8.453(23.094)(63.094)\\
      &\approx 12,316.805

      You could also try verifying that this is the absolute maximum by finding $V''(x)$, but it isn't necessary because (as stated before) $x^3$ is the highest degree. That means only one maximum and one minimum (which we found using the quadratic formula from above).

      ## Conclusion
      To conclude, the most efficient way to fold the $40 \text{ cm} \times 80 \text{ cm}$ is to fold each corner in $8.453 \text{ cm}$. Doing so gives you $\approx 12,316.805 \text{ cm}^3$.

      Alternatively, you could just go onto Desmos and plug in equation from the beginning to find the maximum value.
      (2 votes)
  • spunky sam blue style avatar for user Jai Sankar
    Instead of doing the second derivative test, couldn't you have just plugged in the x values into the original equation?
    (5 votes)
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  • leaf green style avatar for user Ashby Daugherty
    So, what units would this problem be answered in? Would it be unitless, or would you say "units cubed?"
    (4 votes)
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  • winston default style avatar for user Emma Gao
    How does Khan simplify 4^3 - 100x^2 + 600x to 12x^2 - 200x + 600? I am very new to calculus and just came to watch these videos for optimization, so please explain very clearly.
    (5 votes)
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  • leaf green style avatar for user Cat Umpherville
    I'm having problems understanding what I'm looking for. I don't quite fully understand the concept of optimization in total. I have a test today in math and I'm scared I'm not going to do well. Could you please help me understand this? Thank you :)
    (2 votes)
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    • blobby green style avatar for user Creeksider
      I don’t know how much help you can get from a brief reply here, but I’ll offer some comments for what they’re worth.

      Optimization problems have to do with finding a tipping point. Something is getting better up to a point, and then it starts to get worse. It’s getting bigger, then it starts to get smaller. Or it’s getting smaller, then it starts to get bigger.

      We find those tipping points by looking at the derivative, which is the rate at which something is changing. As long as the rate of change is in the “good” direction (which may be up or down, depending on what you’re optimizing), we keep going. The tipping point is where the derivative starts to go in the other direction. It’s the top of a frown-shaped curve or the bottom of a smile-shaped curve. The rate of change at these exact moments is zero, so we hunt for optimization points by finding the derivative and then determining where the derivative is equal to zero (the “critical points”).

      You need to be aware that some functions will have two or more critical points, and you have to use other tests to determine which one is optimal in terms of the way the problem was set forth. Some functions have critical points that aren’t turning points (for example, y = x^3 has a critical point at x = 0 but doesn’t turn down). And some teachers want to test your understanding by giving you a problem where there’s only one critical point, which is a true turning point, but because of the way the problem was stated, this is a worst solution instead of an optimal one.

      Good luck!
      (9 votes)
  • blobby green style avatar for user MSylva1990
    how would you find the dimensions of an open top box with a minimum surface area and a volume of 32000 cubic cm, I'm guessing partial derivatives but lost other than that
    (2 votes)
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    • blobby green style avatar for user Creeksider
      In theory you have three variables to work with: height, width and depth. In practice, it's fairly obvious that the box has to be square, so you have only two variables, one for the height of the box and one for the length of a side (width and depth). If you're expected to prove this part, you begin by showing that the rectangle with minimum perimeter for a fixed area is a square.

      With the problem reduced to height and length of a side, you can create a formula for the surface area (remembering to include the bottom of the box) and set it equal to 32,000. Differentiate and find the place where the derivative is equal to zero, then confirm that this is a minimum.
      (7 votes)
  • purple pi purple style avatar for user Raghav
    In maxima-minima word problems, if something is not given as a constant, and is not mentioned at all, can we take it to be variable?
    (4 votes)
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    • mr pants teal style avatar for user Anton Fagerberg
      This question is tough to answer because it is not clear exactly what kind of situation you're in.

      The word problem could be a general problem, meaning that you should solve the problem for any imaginable number, which gives you an answer, in this case a formula for solving that type of problem.

      If it's a specific problem, you can see the previous video on the graphical solution to find out how Sal managed to find an expression for the different lengths:

      Basically - if you want to find an exact solution you need to be given enough variables to be able to compute the exact answer in the end.

      That does not mean these things are written down for you, it might be implicit in the problem itself, and the problem might be constructed for you to be able to deduce the necessary equation that allows you to find the maximum/minimum of the function

      Hope that helps!
      (2 votes)
  • blobby green style avatar for user Michelle Leung
    isnt x also not equal to 15 because of the polynomial (30-2x)?

    If x = 15 then (30-2x)=0 and then V(x) will also be equal to 0
    (3 votes)
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Video transcript

In the last video, we were able to get a pretty good sense about how large of an x we should cut out of each corner in order to maximize our volume. And we did this graphically. What I want to do in this video is use some of our calculus tools to see if we can come up with the same or maybe even a better result. So to do that, I'm going to have to figure out the critical points of our volume as a function of x. And to do that, I need to take the derivative of the volume. So let me do that. And before I even do that, it'll simplify things so I don't have to use the product rule in some way and then have to simplify that, let me just multiply this expression out. So let's rewrite volume as a function of x is equal to-- and I'll write it all in yellow. So it's going to be x times-- I'll multiply these two binomials first. So 20 times 30 is 600. Then I have 20 times negative 2x, which is negative 40x. Then I have negative 2x times 30, which is negative 60x. And then I have negative 2x times negative 2x, which is positive 4x squared. So this part over here simplifies and I can change the order to 4x squared minus 100x plus 600. I just switched the order in which I'm writing them. So that's that. And so I can rewrite the volume of x as being equal to x times all of this business, which is-- let me make sure I have enough space, let me do it a little higher-- which is equal to 4x to the third power minus 100x squared plus 600x. And that'll be pretty straightforward to take the derivative. So let's say that v prime of x is going to be equal to-- I just have used the power rule multiple times-- so 4 times 3 is 12. x to the 3 minus 1 power 12x squared minus 200 times x to the first power, which is just x, plus 600. And so now we just have to figure out when this is equal to 0. So we have to figure out when 12x squared minus 200x plus 600 is equal to 0. What x values gets my derivative to be equal to 0? When is my slope equal to 0? I could also look for critical points where the derivative is undefined, but this derivative is defined especially throughout my domain of x that I care about, between 0 and 10. So I could try to factor this or try to simplify this a little bit. But I'm just going to cut to the chase and try to use the quadratic formula here. So this is, the x's that satisfy this is going to be x is going to be equal to-- So negative b. So it's 200. Negative, negative 200 is positive 200. Plus or minus the square root of b squared, which is negative 200 squared. So I could just write that as 200 squared. Doesn't matter if it's negative 200 squared or 200 squared, I'm going to get the same value. So let me give myself some more space. So negative 200 squared. Well, that's going to be 4 with 4 0's. 1, 2, 3, 4. So that's going to be 40,000 minus 4ac. So minus 4 times 12, 12 times 600-- I still didn't give myself enough space-- times 600. All of that over 2 times a. So all of that over 24. And I'll take out the calculator again to try to calculate this. So let me get out of graphing mode. All right, so first I'll try when I add the radical. So I'm going to get 200 plus the square root of 40,000. I could have just written that as 200 squared, but that's fine. 40,000 minus 4 times 12 times 600. And I get 305, which I then need to divide by 24. Which I'll divide by 24, and I get 12.74. So one of my possible x's. So it equals 12.74. And now let me do the situation where I subtract what I had in the radical sign. So let me get my calculator back. And so now let me do 200-- I probably could have done this slightly more efficiently, but this is fine-- minus the square root of 40,000-- 1, 2, 3-- minus 4 times 12 times 600. And I get that's just the numerator. And then I'm going to divide that by 24, and I get 3.92. Did I do that right? 200 minus 40,000 minus 4 times 12 times 600, all of that divided by 24. My previous answer divided by 24 gives me 3.92. So it's 12.74 or 3.92. Now which of these can I use? Well x equals 12.74 is outside of our valid values for x. If x was equal to 12.74, the x's would start to overlap with each other. So x cannot be 12.74. So we get a critical point at x is equal to 3.92. And you could look at the graph and you could say, oh, well look, that looks like a maximum value. But if you didn't have the graph at your disposal, you can then do the second derivative test and say, hey, are we concave upwards or concave downwards when x is equal to 3.92? Well, in order to do the second derivative test, you have to figure out the second derivative is. So let's do that. b prime prime of x is going to be equal to 24x, 24 times x to the first, minus 200. And you can just look at inspection that this number right over here is less than 4. So this thing right over here is going to be less than 100. You subtract 200. So we can write the second derivative at 3.92 is going to be less than 0. You can figure out what the exact value is if you like. So because this is less than 0, we are concave downwards. Another way of saying it is the slope is decreasing the entire time. Concave downwards. When the slope is decreasing the entire time, our shape looks like that. The slope could start off high, lower, lower, gets to 0, even lower, lower, lower. And we even saw that on the graph right over here. And since it's concave downwards, that implies that our critical point that sits where the interval is concave downward, that critical point is a local maximum. So this is the x value at which our function attains our maximum. Now what is that maximum value? Well, we could type that back in into our original expression for volume to figure what that is. So let's figure out what the volume when we get to 3.92 is equal to. What is our maximum volume? So get the calculator back out. It's obviously roughly 3.92. I could use this exact value. Actually, I'll just use 3.92 to get a rough sense of what our maximum value is, our maximum volume. So it'll be 3.92. I'll just use this expression for the volume as a function of x. 3.92 times 20 minus 2 times 3.92 times 30 minus 2 times 3.92 gives us-- and we deserve a drum roll now-- gives us 1,056.3. So 1,056.3, which is a higher volume then we got when we just inspected it graphically. We probably could have gotten a little bit more precise if we zoomed in some and then we would have gotten a little bit better of an answer, but there you have it. Analytically, we were able to actually get an even better answer than we were able to do at least on that first pass graphically.